# Time-Stepping Approximation of Rigid-Body Dynamics with Perfect Unilateral Constraints. I: The Inelastic Impact Case

Laetitia Paoli

► To cite this version:

Laetitia Paoli. Time-Stepping Approximation of Rigid-Body Dynamics with Perfect Unilateral Constraints. I: The Inelastic Impact Case. Archive for Rational Mechanics and Analysis, 2010, 198 (2), pp.457-503. 10.1007/s00205-010-0311-0. hal-01566929

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# Time-Stepping Approximation of Rigid-Body Dynamics with Perfect Unilateral Constraints. I: The Inelastic Impact Case

L. PAOLI

We consider a discrete mechanical system with a non-trivial mass matrix, subjected to perfect unilateral constraints described by the geometrical inequalities $f_{\alpha}(q) \ge 0, \alpha \in \{1, \dots, \nu\}(\nu \ge 1)$. We assume that the transmission of the velocities at impact is governed by Newton's Law with a coefficient of restitution $e = 0$ (so that the impact is inelastic). We propose a time-discretization of the second order differential inclusion describing the dynamics, which generalizes the scheme proposed in Paoli (J Differ Equ 211:247-281, 2005) and, for any admissible data, we prove the convergence of approximate motions to a solution of the initial-value problem.

## 1. Introduction

We consider a discrete mechanical system subjected to perfect unilateral constraints. More precisely, let us denote by $u \in \mathbb{R}^d$ the generalized coordinates of a typical configuration of the system. We assume that the set $K$ of admissible configurations is described by $\nu \ge 1$ geometrical inequalities

$$f_{\alpha}(u) \ge 0, \quad \alpha \in \{1, \dots, \nu\}$$

where $f_{\alpha}$ is a smooth function (at least $C^1$) such that $\nabla f_{\alpha}(u)$ does not vanish in a neighbourhood of $\{u \in \mathbb{R}^d; f_{\alpha}(u) = 0\}$.

At each $u \in \mathbb{R}^d$ we define the set of active constraints $J(u)$ by

$$J(u) = \{\alpha \in \{1, \dots, \nu\}; f_{\alpha}(u) \le 0\}.$$

In order to avoid some geometrical inconsistencies we assume, moreover, that the active constraints along $\partial K$ are linearly independent, that is, $(\nabla f_{\alpha}(u))_{\alpha \in J(u)}$ are linearly independent for all $u \in K$.
Then the dynamics is described by the following measure differential inclusion (see [17] or [6] for instance)

$$M(u)\ddot{u} - g(t, u, \dot{u}) \in -N_K(u) \quad (1)$$

where $M(u)$ is the mass matrix of the system and $N_K(u)$ is the normal cone to $K$ at $u$ given by

$$N_K(u) = \begin{cases} \{0\} & \text{if } u \in \text{Int}(K), \\ \left\{ \sum_{\alpha \in J(u)} \lambda_{\alpha} \nabla f_{\alpha}(u), \lambda_{\alpha} \le 0 \; \forall \alpha \in J(u) \right\} & \text{if } u \in \partial K, \\ \emptyset & \text{if } u \notin K. \end{cases}$$

We also define the tangent cone to $K$ at $u$

$$T_K(u) = \{w \in \mathbb{R}^d; (\nabla f_\alpha(u), w) \ge 0 \; \forall \alpha \in J(u)\}$$

where $(v, w)$ denotes the Euclidean scalar product of vectors $v$ and $w$ in $\mathbb{R}^d$. Since $u(s) \in K$ for all $s$, we infer that

$$\dot{u}(t+0) \in T_K(u(t)), \quad \dot{u}(t-0) \in -T_K(u(t)) \quad (t > 0)$$

whenever $\dot{u}(t \pm 0)$ exists. It follows that the velocities are discontinuous at impacts if $\dot{u}(t-0) \notin T_K(u(t))$ and (1) implies that

$$M(u(t)) (\dot{u}(t+0) - \dot{u}(t-0)) \in -N_K(u(t)).$$

This relation does not uniquely determine $\dot{u}(t+0)$, so we should add an impact law. Following MOREAU ([6] and [7], see also [14] or [16]) we assume that

$$\dot{u}(t+0) = \text{Proj}_{M(u(t))}(T_K(u(t)), \dot{u}(t-0)) \quad (2)$$

where $\text{Proj}_{M(u(t))}$ denotes the projection relative to the Riemannian metric defined by the inertia operator $M(u)$.

More precisely, for admissible initial data $(u_0, v_0) \in K \times T_K(u_0)$, we consider the following problem:

**Problem (P)** Find $u : [0, \tau] \to \mathbb{R}^d$ ($\tau > 0$) such that:

(P1) $u$ is an absolutely continuous function from $[0, \tau]$ to $K$ and $\dot{u} \in BV(0, \tau; \mathbb{R}^d)$,

(P2) the differential inclusion

$$M(u)\ddot{u} - g(t, u, \dot{u}) \in -N_K(u)$$

is satisfied in the following sense: there exists a (non-unique) non-negative measure $\mu$ such that the Stieltjes measure $d\dot{u} = \ddot{u}$ and the usual Lebesgue measure $dt$ admit densities with respect to $d\mu$, that is, there exist two $d\mu$-integrable functions $v'_\mu$ and $t'_\mu$ such that $\ddot{u} = d\dot{u} = v'_\mu d\mu$, $dt = t'_\mu d\mu$, and such that

$$M(u(t)) v'_{\mu}(t) - g(t, u(t), \dot{u}(t)) t'_{\mu}(t) \in -N_K(u(t)) d\mu \text{ almost everywhere,} \quad (3)$$
(P3) for all $t \in (0, \tau)$

$$ \dot{u}(t+0) = \operatorname{Proj}_{M(u(t))}(T_K(u(t)), \dot{u}(t-0)) \quad (4) $$

(P4) $u(0) = u_0$, $\dot{u}(0+0) = v_0$.

Observe that the right-hand side of (3) is a cone, so that the differential inclusion remains true for any non-negative measure $\mu$ with respect to which $du$ and $dt$ admit densities (see [7]).

For this model of impact, a very complete theoretical study has been performed by BALLARD in [1]: using existence results for both ordinary differential equations and variational inequalities, he proved the existence and uniqueness of a maximal solution for the initial value problem when the data are analytical. Some counter-examples show that uniqueness may be lost for less regular data (see [7] or [1] for instance) but existence results have still been established in the single constraint case (that is, $v = 1$): see [3,4] and [9,12] for a trivial mass matrix (that is, $M(u) \equiv \mathrm{Id}_{\mathbb{R}^d}$), and [18] and [13,15] for a non-trivial mass matrix. All these results rely on the study of a sequence of approximate solutions constructed either by a penalty method [18] or by a time-stepping scheme [3,4,9,12,13,15].

For the multi-constraint case, these techniques encounter a new difficulty: in general, the motion is not continuous with respect to the data. Nevertheless, some sufficient conditions ensuring continuity on data have been established in [1] and [10]. In this framework, the convergence of the time-stepping scheme proposed in [9] has been extended to the multi-constraint case with inelastic shocks when the mass matrix is trivial and the set $K$ is convex [11]. The aim of this paper is to relax these restricting conditions for the mass matrix and the set $K$, and to prove an analogous convergence result in a more general setting.

More precisely we assume the same kind of regularity for the data as in [10], that is,

(H1) $g$ is a continuous function from $[0, T] \times \mathbb{R}^d \times \mathbb{R}^d$ ($T > 0$) to $\mathbb{R}^d$;

(H2) for all $\alpha \in \{1, \dots, v\}$, the function $f_\alpha$ belongs to $C^1(\mathbb{R}^d)$, $\nabla f_\alpha$ is locally Lipschitz continuous and does not vanish in a neighbourhood of $\{u \in \mathbb{R}^d; f_\alpha(u) = 0\}$;

(H3) the set $K$ is defined by

$$ K = \{u \in \mathbb{R}^d; f_\alpha(u) \ge 0, \alpha \in \{1, \dots, v\}\} $$

and the active constraints along $\partial K$ are functionally independent, that is, the vectors $(\nabla f_\alpha(u))_{J(u)}$ are linearly independent for all $u \in K$;

(H4) $M$ is a mapping of class $C^1$ from $\mathbb{R}^d$ to the set of symmetric positive definite $d \times d$ matrices.

With this last assumption, we may define $M^{-1}(u)$, $M^{1/2}(u)$ and $M^{-1/2}(u)$ for all $u \in \mathbb{R}^d$; the corresponding mappings are of class $C^1$ from $\mathbb{R}^d$ to the set of symmetric positive definite $d \times d$ matrices.
Let $F$ be a function such that

(H5) $F$ is continuous from $[0, T] \times \mathbb{R}^d \times \mathbb{R}^d \times [0, h^*] (h^* > 0)$ to $\mathbb{R}^d$ and is consistent with respect to $g$, that is,

$$F(t, u, v, 0) = M^{-1}(u)g(t, u, v) \quad \forall (t, u, v) \in [0, T] \times \mathbb{R}^d \times \mathbb{R}^d.$$

For admissible initial data $(u_0, v_0) \in K \times T_K(u_0)$, we consider the initial-value problem (P) and we define a time-stepping scheme as follows:

• the initial positions $U^0$ and $U^1$ are given by

$$U^0 = u_0, \quad U^1 \in \operatorname{Argmin}_{Z \in K} \|u_0 + h v_0 + h z(h) - Z\|_{M(u_0)} \quad (5)$$

with $\lim_{h \to 0} z(h) = 0$,

• for all $n \in \{1, \dots, \lfloor T \rfloor\}$, let

$$W^n = 2U^n - U^{n-1} + h^2 F^n, \quad F^n = F\left(nh, U^n, \frac{U^n - U^{n-1}}{h}, h\right) \quad (6)$$

and

$$U^{n+1} \in \operatorname{Argmin}_{Z \in K} \|W^n - Z\|_{M(U^n)} \quad (7)$$

where $\| \cdot \|_{M(U)}$ is the norm associated to the kinetic metric at $U$ defined by $\|Z\|_{M(U)}^2 = (Z, Z)_{M(U)}$ with

$$(Z, Z')_{M(U)} = (Z, M(U)Z') = (M(U)Z, Z')$$

for all $(U, Z, Z') \in (\mathbb{R}^d)^3$.

In the initialization procedure given at formula (5), the mapping $h \mapsto z(h)$ can be chosen in such a way that the unconstrained dynamics is approximated at order $p$, with $p \ge 1$, by $\tilde{U}^1 = u_0 + h v_0 + h z(h)$ at $t_1 = h$. For instance, the simplest choice $z(h) \equiv 0$ leads to $p = 1$, while $z(h) = \frac{h}{2} M^{-1}(u_0)g(0, u_0, v_0)$ leads to $p = 2$. Moreover we can observe that we obtain $U^{n+1} = W^n$ when $W^n \in K$ and thus

$$\frac{U^{n+1} - 2U^n + U^{n-1}}{h^2} = F^n \quad \text{when } W^n \in K,$$

which is a centered time-discretization of the unconstrained dynamics. Furthermore, if $M(u) = \text{Id}_{\mathbb{R}^d}$ for all $u \in \mathbb{R}^d$ and $K$ is convex, we recognize the scheme introduced in [9] for the first time and whose convergence has been established in [9] when $\partial K$ is smooth, and in [11] in the general case.

We now define the approximate solutions $u_h$ by

$$u_h(t) = U^n + (t - nh) \frac{U^{n+1} - U^n}{h} \quad \forall t \in [nh, (n+1)h] \cap [0, T]$$

for all $n \in \{0, \dots, \lfloor T/h \rfloor\}$ and $h \in (0, h^*]$.
Since the impact law (2) leads to some discontinuity with respect to the data if the active constraints at impacts create an obtuse angle (see [10]), we cannot expect convergence of the approximate motions unless we add some assumptions on the geometry of active constraints along $\partial K$.

So, for all $u \in K$ and $\alpha \in J(u)$, let us define

$$e_{\alpha}(u) = \frac{M^{-1/2}(u)\nabla f_{\alpha}(u)}{|M^{-1/2}(u)\nabla f_{\alpha}(u)|}$$

where $|\cdot|$ denotes the Euclidean norm in $\mathbb{R}^d$, and assume that the “angle condition” given in [10], which ensures continuity on data in the case of inelastic shocks, holds. That is,

(H6) for all $u \in \partial K$, for all $(\alpha, \beta) \in J(u)^2$, such that $\alpha \neq \beta$, we have

$$(e_{\alpha}(u), e_{\beta}(u)) \leq 0.$$

This inequality can be interpreted geometrically: in the local momentum metric, defined by the matrix $M^{-1}(u)$, the active constraints create right or acute angles.

Then, under assumptions (H1)–(H6) we prove the convergence of a subsequence of the approximate solutions $(u_h)_{h^* \geq h > 0}$ to a solution of problem (P).

The paper is organized as follows. In the next sections we establish a priori estimates for the discrete velocities and accelerations on a non-trivial time interval $[0, \tau]$, with $0 < \tau \leq T$. Then we pass to the limit when $h$ tends to zero on $[0, \tau]$: using Ascoli’s and Helly’s theorems we obtain the convergence of a subsequence of $(u_h)_{h^* \geq h > 0}$ to a limit $u$ which satisfies (P1) and (P2). Next, for any instant $t$ such that $u(t) \in \partial K$, we perform a precise local study of the approximate motions and we prove that the limit $u$ also satisfies (P3) and (P4). Finally, we conclude the proof with some energy estimates which allow us to obtain global results.

## 2. A priori estimates for the discrete velocities

Let us begin with a priori estimates for the discrete velocities. Let $\mathcal{B}$ be a given convex compact subset of $\mathbb{R}^d$ such that $\mathcal{B} \cap K \neq \emptyset$. Possibly decreasing $h^*$, we may assume without loss of generality that

$$|z(h)| \leq 1 \quad \forall h \in (0, h^*].$$

In this section we consider a more general scheme for which the initialization procedure involves an initial time $t_{0h} \in [0, T)$ depending on $h$. This modification will allow us, in the last section of the paper, to extend the a priori estimates of the discrete velocities by considering as “new” initial data the already constructed approximate positions at some time steps $t_{0h}$ and $t_{0h} + h$.

So, let $h \in (0, h^*]$, $t_{0h} \in [0, T)$, $U^0$ and $U^1$ be given in $\mathcal{B} \cap K$ and $K$ respectively, and for all $n \in \{1, \dots, \lfloor \frac{T-t_{0h}}{h} \rfloor\}$

$$U^{n+1} \in \operatorname{Argmin}_{Z \in K} \|W^n - Z\|_{M(U^n)}$$
with

$$W^n = 2U^n - U^{n-1} + h^2 F^n, \quad F^n = F\left(t_{0h} + nh, U^n, \frac{U^n - U^{n-1}}{h}, h\right).$$

For all $h \in (0, h^*]$ and $n \in \{0, \dots, \lfloor \frac{T-t_{0h}}{h} \rfloor\}$, we define

$$V^n = \frac{U^{n+1} - U^n}{h}.$$

First let us observe that

**Lemma 1.** For all $h \in (0, h^*]$ and $n \in \{1, \dots, \lfloor \frac{T-t_{0h}}{h} \rfloor\}$, we have

$$M(U^n)(V^{n-1} - V^n + hF^n) \in N_K(U^{n+1}).$$

**Proof.** Let $h \in (0, h^*]$ and $n \in \{1, \dots, \lfloor \frac{T-t_{0h}}{h} \rfloor\}$. By definition of $U^{n+1}$ we have

$$\begin{aligned} \|W^n - U^{n+1}\|_{M(U^n)}^2 &\le \|W^n - Z\|_{M(U^n)}^2 \\ &\le \|W^n - U^{n+1}\|_{M(U^n)}^2 + 2(W^n - U^{n+1}, U^{n+1} - Z)_{M(U^n)} \\ &\quad + \|U^{n+1} - Z\|_{M(U^n)}^2 \end{aligned}$$

for all $Z \in K$, which yields

$$(W^n - U^{n+1}, Z - U^{n+1})_{M(U^n)} \le \frac{1}{2} \|U^{n+1} - Z\|_{M(U^n)}^2 \quad \forall Z \in K. \quad (8)$$

If $U^{n+1} \in \text{Int}(K)$, we deduce from (8) that

$$W^n - U^{n+1} = h(V^{n-1} - V^n + hF^n) = 0$$

and

$$M(U^n)(V^{n-1} - V^n + hF^n) \in N_K(U^{n+1}) = \{0\}.$$

Assume now that $U^{n+1} \in \partial K$ and let

$$\tilde{T}_K(U^{n+1}) = \{w \in \mathbb{R}^d; (\nabla f_\alpha(U^{n+1}), w) > 0 \quad \forall \alpha \in J(U^{n+1})\}.$$

For all $w \in \tilde{T}_K(U^{n+1})$ there exists a smooth curve $t \mapsto \varphi(t)$ such that $\varphi(0) = U^{n+1}$, $\varphi'(0) = w$ and $\varphi(t) \in K$ for all $t$ in a right neighbourhood of 0. By choosing $Z = \varphi(t)$ we infer that

$$(W^n - U^{n+1}, w)_{M(U^n)} \le 0 \quad \forall w \in \tilde{T}_K(U^{n+1}).$$

Then the density of $\tilde{T}_K(U^{n+1})$ in $T_K(U^{n+1})$ leads to

$$(V^{n-1} - V^n + hF^n, w)_{M(U^n)} \le 0 \quad \forall w \in T_K(U^{n+1}).$$

Finally, observing that $T_K(u) = N_K(u)^\perp$ for all $u \in K$, we are able to conclude the proof. $\square$
Let us introduce some notation. We define

$$ \lambda_{\max}(u) = \|M(u)\|, \quad \lambda_{\min}(u) = \frac{1}{\|M^{-1}(u)\|} \quad \forall u \in \mathbb{R}^d. $$

Since $u \mapsto M(u)$ is continuous with values in the set of symmetric positive definite matrices, the mappings $u \mapsto \lambda_{\max}(u)$ and $u \mapsto \lambda_{\min}(u)$ are well defined and continuous from $\mathbb{R}^d$ to $\mathbb{R}_+^d$. Moreover

$$ \lambda_{\min}(u)|w|^2 \le \|w\|_{M(u)}^2 \le \lambda_{\max}(u)|w|^2 \quad \forall w \in \mathbb{R}^d, \quad \forall u \in \mathbb{R}^d. $$

Since $\mathcal{B}$ is compact, there exists $\delta > 0$ such that, for all $(q, q') \in \mathcal{B} \times \mathbb{R}^d$ such that $|q - q'| \le \delta$, we have:

$$ |\lambda_{\min}(q) - \lambda_{\min}(q')| \le \frac{1}{2} \inf_{u \in \mathcal{B}} \lambda_{\min}(u), $$

$$ |\lambda_{\max}(q) - \lambda_{\max}(q')| \le \frac{1}{2} \sup_{u \in \mathcal{B}} \lambda_{\max}(u). $$

We define

$$ B_0 = \{u \in \mathbb{R}^d; \operatorname{dist}(u, \mathcal{B}) \le \delta\}. \quad (9) $$

Then $B_0$ is also a convex compact subset of $\mathbb{R}^d$ and we have

$$ \frac{1}{2} \inf_{u \in \mathcal{B}} \lambda_{\min}(u) \le \inf_{u \in B_0} \lambda_{\min}(u), \quad \sup_{u \in B_0} \lambda_{\max}(u) \le \frac{3}{2} \sup_{u \in \mathcal{B}} \lambda_{\max}(u). $$

We let

$$ \lambda_{\min} = \frac{1}{2} \inf_{u \in \mathcal{B}} \lambda_{\min}(u), \quad \lambda_{\max} = \frac{3}{2} \sup_{u \in \mathcal{B}} \lambda_{\max}(u). \quad (10) $$

Of course we have

$$ 0 < \lambda_{\min}|w|^2 \le \|w\|_{M(u)}^2 \le \lambda_{\max}|w|^2 \quad \forall w \in \mathbb{R}^d \setminus \{0\}, \quad \forall u \in B_0. $$

Let $C_0 > 0$ and $C_F$ be given by

$$ C_F = \sup \{|F(t, u, v, h)|; t \in [0, T], u \in B_0 \cup B_1, |v| \le C_0, h \in [0, h^*]\}, \quad (11) $$

where $B_1 = \overline{B}(u_0, C_0T + 1)$. Since the mappings $M, M^{-1}, M^{1/2}$ and $M^{-1/2}$ are of class $C^1$ on $\mathbb{R}^d$, they are Lipschitz continuous on $B_0 \cup B_1$ and we denote by $L_M, L_{M^{-1}}, L_{M^{1/2}}$ and $L_{M^{-1/2}}$ the corresponding Lipschitz constants. Moreover, the functions $\nabla f_\alpha$, $1 \le \alpha \le \nu$, are locally Lipschitzian and there exists also a positive real number $L_f$ such that

$$ |\nabla f_{\alpha}(Z) - \nabla f_{\alpha}(Z')| \le L_f |Z - Z'| \quad \forall (Z, Z') \in (B_0 \cup B_1)^2, \quad \forall \alpha \in \{1, \dots, \nu\}. $$

Next, we obtain some rough estimates on the discrete velocities. More precisely, let us assume that

$$ |V^l| \le C_0 \quad \forall l \in \{0, \dots, n-1\} $$

for some $n \ge 1$. We obtain the following estimate on $V^n$:
**Proposition 1.** Let $C_0 > 0$ and $h_0^* \in (0, h^*]$ such that

$$h_0^* \leq \min\left(\frac{C_0}{C_F}, \frac{\delta}{8C_0}\sqrt{\frac{\lambda_{\min}}{\lambda_{\max}}}\right)$$

where $C_F$ is defined by (11), and $\lambda_{\min}, \lambda_{\max}$ are defined by (10). Let $h \in (0, h_0^*]$, $\tau_h = \min(\delta/(2C_0), T - t_{0h})$ and assume that there exists $n \in \{1, \dots, \lfloor \tau_h/h \rfloor\}$ such that

$$|V^l| \leq C_0 \quad \forall l \in \{0, \dots, n-1\}.$$

Then

$$|V^n| \leq 4 \sqrt{\frac{\lambda_{\max}}{\lambda_{\min}}} C_0.$$

Moreover, for all $l \in \{0, \dots, n\}$ such that $J(U^{l+1}) \neq \emptyset$,

$$(\nabla f_{\alpha}(U^{l+1}), V^l) \leq \frac{L_f h}{2} |V^l|^2 \quad \forall \alpha \in J(U^{l+1}).$$

**Proof.** For all $l \in \{0, \dots, n\}$ we have $U^l \in B_0$, since

$$|U^l - U^0| \leq \sum_{k=0}^{l-1} h |V^k| \leq l h C_0 \leq \tau_h C_0 \leq \delta \quad \forall l \in \{0, \dots, n\}.$$

By definition of $U^{n+1}$ we have

$$\|W^n - U^{n+1}\|_{M(U^n)} \leq \|W^n - U^n\|_{M(U^n)}$$

since $U^n \in K$ and

$$\begin{aligned} W^n - U^{n+1} &= h(V^{n-1} - V^n + hF^n), \\ W^n - U^n &= h(V^{n-1} + hF^n). \end{aligned}$$

Hence

$$\|V^n\|_{M(U^n)} \leq 2\|V^{n-1}\|_{M(U^n)} + 2h\|F^n\|_{M(U^n)}$$

and

$$|V^n| \leq \sqrt{\frac{\lambda_{\max}}{\lambda_{\min}}} (2C_0 + 2hC_F) \leq 4\sqrt{\frac{\lambda_{\max}}{\lambda_{\min}}} C_0.$$

Now, we infer that $U^{n+1} \in B_0$. Indeed

$$|U^{n+1} - U^0| \leq |U^n - U^0| + h|V^n| \leq C_0 \tau_h + 4h \sqrt{\frac{\lambda_{\max}}{\lambda_{\min}}} C_0 \leq C_0 \tau_h + \frac{\delta}{2} \leq \delta.$$
Let $l \in \{0, \dots, n\}$ such that $J(U^{l+1}) \neq \emptyset$. For all $\alpha \in J(U^{l+1})$, we have

$$0 \le f_{\alpha}(U^l) - f_{\alpha}(U^{l+1}) = \int_0^1 (\nabla f_{\alpha}(U^{l+1} + t(U^l - U^{l+1})), U^l - U^{l+1}) \, dt$$

and thus

$$\begin{aligned} (\nabla f_{\alpha}(U^{l+1}), V^l) &\le -\int_0^1 (\nabla f_{\alpha}(U^{l+1} + t(U^l - U^{l+1}))) - \nabla f_{\alpha}(U^{l+1}), V^l \, dt \\ &\le \int_0^1 |\nabla f_{\alpha}(U^{l+1} + t(U^l - U^{l+1})) - \nabla f_{\alpha}(U^{l+1})| |V^l| \, dt. \end{aligned}$$

It follows that

$$(\nabla f_{\alpha}(U^{l+1}), V^l) \le \frac{L_f h}{2} |V^l|^2.$$

□

Now we prove a more precise estimate on the discrete velocities. We have the following result:

**Proposition 2.** Let $C_0 > 0$ and assume that there exist $C_0^* > 0$ and $h_0^* \in (0, h^*]$ such that

$$\begin{aligned} h_0^* &\le \min\left(\frac{C_0}{C_F}, \frac{\delta}{8C_0}\sqrt{\frac{\lambda_{\min}}{\lambda_{\max}}}\right), \\ |V^0| &= \left|\frac{U^1 - U^0}{h}\right| \le C_0^* < \sqrt{\frac{\lambda_{\min}}{\lambda_{\max}}} C_0 \quad \forall h \in (0, h_0^*], \end{aligned}$$

where $C_F$ is defined by (11) and $\lambda_{\min}, \lambda_{\max}$ are defined by (10). Then, there exists $\tau_0 > 0$, depending only on $B, C_0, C_0^*$ and the data, such that

$$|V^n| = \left| \frac{U^{n+1} - U^n}{h} \right| \le C_0 \quad \forall nh \in [0, \min(\tau_0, T - t_{0h})), \quad \forall h \in (0, h_0^*].$$

**Proof.** Let us assume that $h \in (0, h_0^*]$ and $n \in \{1, \dots, \lfloor \tau_h/h \rfloor\}$ such that

$$|V^l| \le C_0 \quad \forall l \in \{0, \dots, n-1\}$$

with $\tau_h = \min(\delta/(2C_0), T - t_{0h})$. Then, with Proposition 1, we know that $U^l \in B_0$ for all $l \in \{0, \dots, n+1\}$ and

$$|V^n| \le 4 \sqrt{\frac{\lambda_{\max}}{\lambda_{\min}}} C_0.$$

Moreover, from Lemma 1 we know that

$$M(U^n) (V^{n-1} - V^n + hF^n) \in N_K(U^{n+1}).$$
It follows that

$$V^n = V^{n-1} + hF^n \quad \text{if } J(U^{n+1}) = \emptyset.$$

Let us assume now that $J(U^{n+1}) \neq \emptyset$.

Using once again, Proposition 1, for all $\alpha \in J(U^{n+1})$ we have also

$$(\nabla f_{\alpha}(U^{n+1}), V^n) \leq \frac{L_f h}{2} |V^n|^2$$

and thus, with Lemma 7 (see Appendix),

$$(e_{\alpha}(U^{n+1}), M^{1/2}(U^{n+1})V^n) \leq \frac{L_f h}{2} \frac{|V^n|^2}{|M^{-1/2}(U^{n+1})\nabla f_{\alpha}(U^{n+1})|} \leq \frac{L_f h}{2m_{B_0}} |V^n|^2.$$

For sake of simplicity, denote $J_n = J(U^{n+1})$ and $e_{\alpha}^n = e_{\alpha}(U^{n+1})$ for all $\alpha \in J_n$. From assumption (H3) we know that $(\nabla f_{\alpha}(U^{n+1}))_{\alpha \in J_n}$ is linearly independent. Thus $(e_{\alpha}^n)_{\alpha \in J_n}$ is also linearly independent and, using Lemma 7, we know that there exist two dual bases $(v_j(U^{n+1}))_{1 \leq j \leq d}$ and $(w_j(U^{n+1}))_{1 \leq j \leq d}$ such that

$$|v_j(U^{n+1})| = 1 \quad \forall j \in \{1, \dots, d\}, \quad v_j(U^{n+1}) = e_j^n \quad \forall j \in J_n$$

and

$$|w_j(U^{n+1})| \leq C_{*,B_0} \quad \forall j \in \{1, \dots, d\}$$

where $C_{*,B_0}$ depends only on the compact set $B_0$ and the mappings $f_{\alpha}$, $\alpha \in \{1, \dots, v\}$, and $M$.

Now, we define

$$w^n = -V^n + \frac{L_f h}{2m_{B_0}} |V^n|^2 \sum_{\alpha \in J_n} M^{-1/2}(U^{n+1}) w_{\alpha}(U^{n+1}).$$

For all $\alpha \in J_n$ we have clearly

$$(\nabla f_{\alpha}(U^{n+1}), -w^n) = \left| M^{-1/2}(U^{n+1}) \nabla f_{\alpha}(U^{n+1}) \right| \left( \left( e_{\alpha}^n, M^{1/2}(U^{n+1})V^n \right) - \frac{L_f h}{2m_{B_0}} |V^n|^2 \right) \leq 0,$$

that is, $w^n \in T_K(U^{n+1})$. With Lemma 1 we get

$$(V^{n-1} - V^n + hF^n, w^n)_{M(U^n)} \leq 0$$

which yields

$$\left( V^n - V^{n-1} - hF^n, V^n - \frac{L_f h}{2m_{B_0}} |V^n|^2 \sum_{\alpha \in J_n} M^{-1/2}(U^{n+1}) w_{\alpha}(U^{n+1}) \right)_{M(U^n)} \leq 0. \qquad (12)$$
It follows that

$$
\begin{align*}
\|V^n\|_{M(U^n)}^2 &\leq -\frac{L_f h}{2m_{B_0}} |V^n|^2 \sum_{\alpha \in J_n} (M^{-1/2}(U^{n+1})w_\alpha(U^{n+1}), V^{n-1}-V^n+hF^n)_{M(U^n)} \\
&\quad + (V^n, V^{n-1}+hF^n)_{M(U^n)} \leq \|V^n\|_{M(U^n)} \|V^{n-1}+hF^n\|_{M(U^n)} \\
&\quad + \frac{L_f h}{2m_{B_0}} \frac{\lambda_{\max}}{\lambda_{\min}} |V^n| \|V^n\|_{M(U^n)} \sum_{\alpha \in J_n} |w_\alpha(U^{n+1})| |V^{n-1}-V^n+hF^n|.
\end{align*}
$$

Using Proposition 1 we get

$$
\begin{align*}
|V^n| \sum_{\alpha \in J_n} |w_\alpha(U^{n+1})| |V^{n-1} - V^n + hF^n| \\
\le 4 \sqrt{\frac{\lambda_{\max}}{\lambda_{\min}}} \nu C_{0,C_{*},B_0} \left( C_0 \left( 1 + 4 \sqrt{\frac{\lambda_{\max}}{\lambda_{\min}}} \right) + hC_F \right)
\end{align*}
$$

if $J_n \neq \emptyset$. Recalling that

$$
V^n = V^{n-1} + hF^n \quad \text{if } J_n = \emptyset
$$

we obtain finally that

$$
\|V^n\|_{M(U^n)} \le \|V^{n-1}\|_{M(U^{n-1})} + h\sqrt{\lambda_{\max}} C_F \\
\qquad + \frac{2L_f h}{m_{B_0}} \left(\frac{\lambda_{\max}}{\lambda_{\min}}\right)^{3/2} \nu C_{*,B_0} C_0^2 \left(2 + 4\sqrt{\frac{\lambda_{\max}}{\lambda_{\min}}}\right)
$$

whenever $J_n = \emptyset$ or $J_n \neq \emptyset$.

By using the Lipschitz property of $M^{1/2}$ on $B_0$, we also have

$$
\begin{align*}
\|V^{n-1}\|_{M(U^n)} &\le \|V^{n-1}\|_{M(U^{n-1})} + \|M^{1/2}(U^n) - M^{1/2}(U^{n-1})\| |V^{n-1}| \\
&\le \|V^{n-1}\|_{M(U^{n-1})} + L_{M^{1/2}h} |V^{n-1}|^2.
\end{align*}
$$

We infer that

$$
\|V^n\|_{M(U^n)} \le \|V^{n-1}\|_{M(U^{n-1})} + C_1 h \quad (13)
$$

with

$$
C_1 = \sqrt{\lambda_{\max}} C_F + L_{M^{1/2}} C_0^2 + \frac{2L_f}{m_{B_0}} \left( \frac{\lambda_{\max}}{\lambda_{\min}} \right)^{3/2} \nu C_{*,B_0} C_0^2 \left( 2 + 4 \sqrt{\frac{\lambda_{\max}}{\lambda_{\min}}} \right).
$$

It follows that

$$
|V^n| \le \sqrt{\frac{\lambda_{\max}}{\lambda_{\min}}} |V^0| + \frac{nC_1h}{\sqrt{\lambda_{\min}}}.
$$
Then, choosing $\tau_0 \in (0, \delta/(2C_0)]$ such that

$$ \sqrt{\frac{\lambda_{\max}}{\lambda_{\min}}} C_0^* + \frac{\tau_0 C_1}{\sqrt{\lambda_{\min}}} \le C_0 $$

and observing that

$$ |V^0| \le C_0^* < \sqrt{\frac{\lambda_{\min}}{\lambda_{\max}}} C_0 \le C_0 \quad \forall h \in (0, h_0^*] $$

we may conclude the proof by induction on $n$. $\square$

Let us now consider the initialization procedure given by formula (5), that is, let $t_{0h} = 0$ and $U^0$ and $U^1$ be given by

$$ U^0 = u_0, \quad U^1 \in \operatorname{Argmin}_{Z \in K} \|u_0 + hv_0 + hz(h) - Z\|_{M(u_0)}, \quad \lim_{h \to 0} z(h) = 0, $$

for all $h \in (0, h^*]$. We can choose $B = \bar{B}(u_0, C+1)$ with $C \ge 0$, and the previous results lead to a uniform estimate of the discrete velocities on a non-trivial time interval. More precisely, we obtain

**Theorem 1.** For all $C_0^* \ge 2\sqrt{\frac{\lambda_{\max}}{\lambda_{\min}}}(|v_0|+1)$ and for all $C_0 > \sqrt{\frac{\lambda_{\max}}{\lambda_{\min}}}C_0^*$, there exist $h_0^* \in (0, h^*]$ and $\tau_0 > 0$, depending only on $B$, $C_0$, $C_0^*$ and the data, such that

$$ |V^n| = \left| \frac{U^{n+1} - U^n}{h} \right| \le C_0 \quad \forall nh \in [0, \min(\tau_0, T)], \quad \forall h \in (0, h_0^*]. $$

**Proof.** Let $C_0^* \ge 2\sqrt{\frac{\lambda_{\max}}{\lambda_{\min}}}(|v_0|+1)$, $C_0 > \sqrt{\frac{\lambda_{\max}}{\lambda_{\min}}}C_0^*$ and $h_0^* \in (0, h^*]$ such that

$$ h_0^* \le \min \left( \frac{C_0}{C_F}, \frac{\delta}{8C_0} \sqrt{\frac{\lambda_{\min}}{\lambda_{\max}}} \right) $$

where $C_F$ is defined by (11) and $\lambda_{\min}$, $\lambda_{\max}$ are defined by (10).

By definition of $U^1$, we have

$$ \|u_0 + hv_0 + hz(h) - U^1\|_{M(u_0)} \le \|u_0 + hv_0 + hz(h) - Z\|_{M(u_0)} \quad \forall Z \in K $$

and by choosing $Z = u_0 = U^0$ we get

$$ \|V^0\|_{M(U^0)} = \left\| \frac{U^1 - U^0}{h} \right\|_{M(u_0)} \le 2 \|v_0 + z(h)\|_{M(u_0)}. $$

Thus,

$$ |V^0| \le 2\sqrt{\frac{\lambda_{\max}}{\lambda_{\min}}} |v_0 + z(h)| \le 2\sqrt{\frac{\lambda_{\max}}{\lambda_{\min}}} (|v_0| + 1) \quad \forall h \in (0, h^*] $$

and

$$ |V^0| \le C_0^* < \sqrt{\frac{\lambda_{\min}}{\lambda_{\max}}} C_0 \quad \forall h \in (0, h_0^*]. $$

It follows that we may apply Proposition 2, which yields the announced result. $\square$
### 3. Convergence of the approximate solutions $(u_h)_{h^* \ge h > 0}$

Before passing to the limit as $h$ tends to zero in the sequence $(u_h)_{h^* \ge h > 0}$, we prove an estimate for the discrete accelerations.

**Proposition 3.** Let us now assume that there exist $C_0 > 0$, $\tau_0 > 0$, $h_0^* \in (0, h^*]$ and a sequence $(h_i)_{i \in \mathbb{N}}$, decreasing to zero, such that

$$ |V^n| \le C_0 \quad \forall nh_i \in [0, \min(\tau_0, T)], \quad \forall h_i \in (0, h_0^*]. \quad (14) $$

Then there exist $h_1^* \in (0, h_0^*]$ and $C_0' > 0$ such that, for all $h_i \in (0, h_1^*]$

$$ \sum_{n=1}^{N} |V^n - V^{n-1}| \le C_0' \quad \text{with} \quad N = \left\lfloor \frac{\min(\tau_0, T)}{h_i} \right\rfloor. $$

**Proof.** The main ideas of the proof are the same as in proposition 2.4 in [11]. More precisely, let $\mathcal{B} = \bar{B}(u_0, C + 1)$ with $C \ge 0$, $B_0$ be defined by (9) and $C_F$ be defined by (11). Without loss of generality, possibly decreasing $h_0^*$, we assume that $C_0 h_0^* \le 1$ and $C_F h_0^* \le C_0$. We denote $K_1 = K \cap B_1 = K \cap \bar{B}(u_0, C_0 T + 1)$ and

$$ \lambda_{\min, B_1} = \inf_{u \in B_1} \lambda_{\min}(u) = \frac{1}{\sup_{u \in B_1} \|M^{-1}(u)\|}, \quad (15) $$

$$ \lambda_{\max, B_1} = \sup_{u \in B_1} \lambda_{\max}(u) = \sup_{u \in B_1} \|M(u)\|. $$

Let $h_i \in (0, h_0^*]$. By definition of the scheme, we have $U^n \in K$ for all $n \in \{0, \dots, \lfloor T/h_i \rfloor + 1\}$. Assumption (14) implies that

$$ |U^n - U^0| \le h_i \sum_{k=0}^{n-1} |V^k| \le nh_i C_0 \le C_0 T + 1 \quad \forall n \in \{0, \dots, N+1\} $$

thus $U^n \in K_1$ for all $n \in \{0, \dots, N+1\}$.

By Lemma 8 (see Appendix), we infer that, for all $q \in K_1$, there exist $a_q \in \mathbb{R}^d$ and two strictly positive numbers $\delta_q$ and $r_q$ such that, for all $q' \in \bar{B}(q, 2\delta_q)$

$$ \bar{B}(a_q, r_q) \subset T_K(q'). \quad (16) $$

It is obvious that $K_1 \subset \bigcup_{q \in K_1} B(q, \delta_q)$, and a compactness argument implies that there exists $(q_j)_{1 \le j \le \ell}$ such that $q_j \in K_1$ for all $j \in \{1, \dots, \ell\}$ and

$$ K_1 \subset \bigcup_{j=1}^{\ell} B(q_j, \delta_{q_j}). $$

In the remainder of the proof we will simply write $\delta_j, a_j$ and $r_j$ instead of $\delta_{q_j}, a_{q_j}$ and $r_{q_j}$. We define

$$ r = \min_{1 \le j \le \ell} r_j, \quad \delta' = \min_{1 \le j \le \ell} \delta_j, \quad \tau_1 = \frac{\delta'}{C_0}. $$
Let $h_1^* \in (0, \min(h_0^*, \tau_1/2))$, $h_i \in (0, h_1^*]$ and $n \in \{0, \dots, N\}$. Let $j \in \{1, \dots, \ell\}$ be such that $U^{n+1} \in B(q_j, \delta_j)$. Then, for all $m \in \{n+1, \dots, p\}$ with $p = \min(N, n + \lfloor \tau_1/h_i \rfloor)$, we have

$$ \begin{aligned} |U^{m+1} - q_j| &\le |U^{m+1} - U^{n+1}| + |U^{n+1} - q_j| \\ &\le \left| \sum_{k=n+1}^{m} h_i V^k \right| + \delta_j \le h_i C_0(m-n) + \delta_j \le \delta' + \delta_j \le 2\delta_j. \end{aligned} $$

By applying (16), we obtain that, for all $m \in \{n+1, \dots, p\}$, we have $\bar{B}(a_j, r_j) \subset T_K(U^{m+1})$. Thus

$$ \bar{B}_{M(U^m)}(a_j, \sqrt{\lambda_{\min,B_1} r_j}) = \{z \in \mathbb{R}^d; \|z - a_j\|_{M(U^m)} \le \sqrt{\lambda_{\min,B_1} r_j}\} \subset \bar{B}(a_j, r_j). $$

Then, we use a classical result about contractions on Hilbert spaces due to MOREAU [5] and we infer that, for all $z \in \mathbb{R}^d$

$$ \begin{aligned} \|z - \operatorname{Proj}_{M(U^m)}(T_K(U^{m+1}), z)\|_{M(U^m)} &\le \frac{1}{2\sqrt{\lambda_{\min,B_1} r_j}} \left( \|z - a_j\|_{M(U^m)}^2 - \|\operatorname{Proj}_{M(U^m)}(T_K(U^{m+1}), z) - a_j\|_{M(U^m)}^2 \right). \end{aligned} $$

With Lemma 1 we know that

$$ M(U^m)(V^{m-1} - V^m + h_i F^m) \in N_K(U^{m+1}). $$

Since $N_K(U^{m+1})$ and $T_K(U^{m+1})$ are two closed convex polar cones, we get

$$ \operatorname{Proj}_{M(U^m)}(T_K(U^{m+1}), V^{m-1} - V^m + h_i F^m) = 0. $$

Hence

$$ \begin{aligned} \|V^{m-1} - V^m + h_i F^m\|_{M(U^m)} &\le \frac{1}{2\sqrt{\lambda_{\min,B_1} r_j}} \left( \| (V^{m-1} - V^m + h_i F^m) - a_j \|_{M(U^m)}^2 - \|a_j\|_{M(U^m)}^2 \right) \\ &\le \frac{1}{2\sqrt{\lambda_{\min,B_1} r_j}} \left( \|V^{m-1} - V^m + h_i F^m\|_{M(U^m)}^2 \right. \\ &\qquad \left. - 2(a_j, V^{m-1} - V^m + h_i F^m)_{M(U^m)} \right). \end{aligned} $$

It follows that

$$ \begin{aligned} \|V^{m-1} - V^m\|_{M(U^m)} \le h_i \|F^m\|_{M(U^m)} &+ \frac{1}{2\sqrt{\lambda_{\min,B_1} r_j}} \left( \|V^{m-1} - V^m\|_{M(U^m)}^2 \right. \\ &+ 2h_i (F^m, V^{m-1} - V^m)_{M(U^m)} + h_i^2 \|F^m\|_{M(U^m)}^2 \\ &\left. - 2(a_j, V^{m-1} - V^m + h_i F^m)_{M(U^m)} \right). \end{aligned} \quad (17) $$
If $J_m \neq \emptyset$, we can reproduce the same computations as in Propositions 1 and 2 to obtain (see (12))

$$ \left( V^m - V^{m-1} - h_i F^m, V^m - \frac{L_f h_i}{2m_{B_1}} |V^m|^2 \sum_{\alpha \in J_m} M^{-1/2}(U^{m+1}) w_\alpha(U^{m+1}) \right)_{M(U^m)} \le 0 $$

which yields

$$ \begin{aligned} - \left(V^{m-1}, V^m\right)_{M(U^m)} &\le -\|V^m\|_{M(U^m)}^2 + h_i (F^m, V^m)_{M(U^m)} \\ &\quad + \frac{L_f h_i \lambda_{\max, B_1}}{2m_{B_1} \sqrt{\lambda_{\min, B_1}}} \nu C_{*, B_1} |V^m|^2 |V^{m-1} - V^m + h_i F^m| \end{aligned} \tag{18} $$

where $C_{*,B_1}$ is the constant defined at Lemma 7.

Otherwise, if $J_m = \emptyset$, we have $V^m = V^{m-1} + h_i F^m$, hence

$$ \begin{aligned} -(V^{m-1}, V^m)_{M(U^m)} &= (-V^m + h_i F^m, V^m)_{M(U^m)} \\ &= -\|V^m\|_{M(U^m)}^2 + h_i (F^m, V^m)_{M(U^m)} \end{aligned} $$

and (18) is still true.

Thus

$$ \|V^{m-1} - V^m\|_{M(U^m)}^2 \le \|V^{m-1}\|_{M(U^m)}^2 - \|V^m\|_{M(U^m)}^2 + 2h_i (F^m, V^m)_{M(U^m)} \\ + \frac{L_f h_i \lambda_{\max, B_1}}{m_{B_1} \sqrt{\lambda_{\min, B_1}}} \nu C_{*, B_1} C_0^2 (2C_0 + h_i C_F). $$

Going back to (17) and using the Lipschitz property of $M$ on $B_1$, we obtain

$$ \|V^{m-1} - V^m\|_{M(U^m)} \le h_i C'_1 + \frac{1}{2\sqrt{\lambda_{\min, B_1} r}} \left( \|V^{m-1}\|_{M(U^{m-1})}^2 - \|V^m\|_{M(U^m)}^2 \right) \\ - 2 (a_j, V^{m-1})_{M(U^{m-1})} + 2 (a_j, V^m)_{M(U^m)} \right) $$

for all $m \in \{n+1, \dots, p\}$, where

$$ C'_1 = \sqrt{\lambda_{\max, B_1}} C_F \left( 1 + \frac{\sqrt{\lambda_{\max, B_1}} (C_0 + a)}{\sqrt{\lambda_{\min, B_1}} r} \right) + C_F^2 \frac{h_0^* \lambda_{\max, B_1}}{2\sqrt{\lambda_{\min, B_1}} r} \\ + \frac{C_0 + 2a}{2\sqrt{\lambda_{\min, B_1}} r} C_0^2 L_M + 3 \frac{L_f \lambda_{\max, B_1}}{2rm_{B_1} \lambda_{\min, B_1}} \nu C_{*, B_1} C_0^3 $$

and $a = \max_{1 \le j \le l} |a_j|$. By summation we get

$$ \begin{aligned} \sum_{m=n+1}^{p} \|V^{m-1} - V^m\|_{M(U^m)} \le (p-n)h_i C'_1 \\ + \frac{1}{2\sqrt{\lambda_{\min, B_1} r}} \left( \|V^n\|_{M(U^n)}^2 - \|V^p\|_{M(U^p)}^2 \right) \\ + 4\lambda_{\max, B_1} C_0 a \end{aligned} $$
Recalling that $p = \min(N, n + \lfloor \tau_1 / h_i \rfloor)$, we infer that

$$ \sum_{m=1}^{N} \|V^{m-1} - V^m\|_{M(U^m)} \le N h_i C_1' + \frac{1}{2\sqrt{\lambda_{\min,B_1}r}} \left( \|V^0\|_{M(U^0)}^2 - \|V^N\|_{M(U^N)}^2 \right) \\ + (k_1+1) \frac{2\lambda_{\max,B_1} C_0 a}{\sqrt{\lambda_{\min,B_1}r}} $$

where $k_1 \in \mathbb{N}$ is such that

$$ 1 + k_1 \left\lfloor \frac{\tau_1}{h_i} \right\rfloor \le N < (k_1 + 1) \left\lfloor \frac{\tau_1}{h_i} \right\rfloor. $$

Observing that $k_1 \le \min(\tau_0, T)/(\tau_1 - h_i)$ for all $h_i \in (0, h_1^*)$, and

$$ |V^{m-1} - V^m| \le \frac{1}{\sqrt{\lambda_{\min,B_1}}} \|V^{m-1} - V^m\|_{M(U^m)} \quad \forall m \in \{1, \dots, N\}, $$

we can conclude the proof with

$$ C_0' = \frac{1}{\sqrt{\lambda_{\min,B_1}}} \left( T C_1' + \frac{\lambda_{\max,B_1} C_0^2}{\sqrt{\lambda_{\min,B_1}r}} + \frac{2\lambda_{\max,B_1} C_0 a}{\sqrt{\lambda_{\min,B_1}r}} \left( \frac{T}{\tau_1 - h_1^*} + 1 \right) \right). $$

□

With these results we can now pass to the limit as $h$ tends to zero. Let us recall the definition of the approximate solutions $(u_h)_{h^* \ge h > 0}$:

$$ u_h(t) = U^n + (t - nh) \frac{U^{n+1} - U^n}{h} \quad \forall t \in [nh, (n+1)h] \cap [0, T] \quad (19) $$

and let us define

$$ v_h(t) = V^n = \frac{U^{n+1} - U^n}{h} \quad \forall t \in [nh, (n+1)h] \cap [0, T] \quad (20) $$

for all $n \in \{0, \dots, \lfloor T/h \rfloor\}$ and $h \in (0, h^*)$.

Let us assume from now on that

(H7) there exist $C_0 > 0$, $\tau_0 > 0$, $h_0^* \in (0, h^*)$ and a subsequence $(h_i)_{i \in \mathbb{N}}$, decreasing to zero, such that

$$ |V^n| \le C_0 \quad \forall nh_i \in [0, \min(\tau_0, T)] \quad \forall h_i \in (0, h_0^*]. $$

We define $B = \bar{B}(u_0, C+1)$ with $C \ge 0$. Let $B_0$ and $C_F$ be defined by (9) and (11), respectively. We assume (without loss of generality) that $C_0 h_0^* \le 1$ and $C_F h_0^* \le C_0$. Let us denote $\tau = \min(\tau_0, T)$. From assumption (H7) and Proposition 3 we know that $(u_{h_i})_{h_1^* \ge h_i > 0}$ is uniformly $C_0$-Lipschitz continuous on $[0, \tau]$ and $(v_{h_i})_{h_1^* \ge h_i > 0}$ is uniformly bounded in $L^\infty(0, \tau; \mathbb{R}^d) \cap BV(0, \tau; \mathbb{R}^d)$. It follows that $(u_{h_i})_{h_1^* \ge h_i > 0}$ is equicontinuous and, using Ascoli's and Helly's theorems, we
infer that there exists a subsequence, still denoted $(h_i)_{i \in \mathbb{N}}$, $u \in C^0([0, \tau]; \mathbb{R}^d)$ and $v \in BV(0, \tau; \mathbb{R}^d)$, such that

$$u_{h_i} \to u \quad \text{strongly in } C^0([0, \tau]; \mathbb{R}^d), \qquad (21)$$

and

$$v_{h_i} \to v \quad \text{pointwise in } [0, \tau]. \qquad (22)$$

Moreover, we have

$$u_{h_i}(t) = u_0 + \int_0^t v_{h_i}(s) \, ds \quad \forall t \in [0, T], \quad \forall h_i \in (0, h^*].$$

Thus, with Lebesgue's theorem, we get

$$u(t) = \lim_{h_i \to 0} \left( u_0 + \int_0^t v_{h_i}(s) \, ds \right) = u_0 + \int_0^t v(s) \, ds \quad \forall t \in [0, \tau]. \quad (23)$$

We infer that $u$ is $C_0$-Lipschitz continuous and

$$u_{h_i}(t), u(t) \in \bar{B}(u_0, C_0\tau) \subset B_1 = \bar{B}(u_0, C_0T + 1) \quad \forall t \in [0, \tau], \forall h_i \in (0, h_1^*].$$

Moreover, $u$ is absolutely continuous on $[0, \tau]$, thus $u$ admits a derivative (in the classical sense) almost everywhere on $[0, \tau]$ and $\dot{u} \in L^1(0, \tau; \mathbb{R}^d)$. From (23) we infer that $\dot{u}(t) = v(t)$ for all $t \in [0, \tau]$ such that $v$ is continuous at $t$. Possibly modifying $\dot{u}$ on a countable subset of $[0, \tau]$, we may assume without loss of generality that $\dot{u} = v$.

As usual, we adopt the convention

$$\begin{align}
\dot{u}(0-0) &= v(0-0) = v(0) = \dot{u}(0), \tag{24} \\
\dot{u}(\tau+0) &= v(\tau+0) = v(\tau) = \dot{u}(\tau). \nonumber
\end{align}$$

Then we observe that

**Lemma 2.** For all $t \in [0, \tau]$, $u(t) \in K$.

**Proof.** Let $t \in [0, \tau]$. For all $h_i \in (0, h_1^*]$ there exists $n \in \{0, \dots, \lfloor \frac{\tau}{h_i} \rfloor\}$ such that $t \in [nh_i, (n+1)h_i)$. Then, observing that $U^n \in K$ we get

$$\begin{align*}
\mathrm{dist}(u(t), K) &\le |u(t) - U^n| \le |u(t) - u_{h_i}(t)| + |u_{h_i}(t) - U^n| \\
&\le |u(t) - u_{h_i}(t)| + (t - nh_i)|V^n| \le \|u - u_{h_i}\|_{C^0([0, \tau]; \mathbb{R}^d)} + C_0h_i.
\end{align*}$$

By passing to the limit as $h_i$ tends to zero, we obtain $\mathrm{dist}(u(t), K) \le 0$, that is, $u(t) \in K$. $\square$
### 3.1. Study of property (P2)

Now let us prove that $u$ satisfies property (P2), that is, the differential inclusion (3). First, we observe that there exists at least one non-negative measure $\mu$ such that the Stieltjes measure $\ddot{u} = d\dot{u} = dv$ and the usual Lebesgue measure $dt$ admit densities with respect to $\mu$. Indeed, let $\mu$ be defined by $d\mu = |d\dot{u}| + dt : \mu$ is non-negative and the measures $\ddot{u} = d\dot{u}$ and $dt$ are both absolutely continuous with respect to $\mu$.

Now, let $\mu = |d\dot{u}| + dt$. We denote by $v'_{\mu}$ and $t'_{\mu}$ the densities of $d\dot{u} = dv$ and $dt$ with respect to $d\mu$. We have to prove that

$$M(u(t)) v'_{\mu}(t) - g(t, u(t), \dot{u}(t)) t'_{\mu}(t) \in -N_K(u(t)) \quad d\mu \text{ almost everywhere.}$$

By Jeffery's theorem (see [2] or [4]) we know that there exists a $d\mu$-negligible set $N \subset [0, \tau]$ such that, for all $t \in [0, \tau] \setminus N$:

$$v'_{\mu}(t) = \lim_{\epsilon \to 0^+} \frac{d\dot{u}(I_{\epsilon})}{d\mu(I_{\epsilon})}, \quad t'_{\mu}(t) = \lim_{\epsilon \to 0^+} \frac{dt(I_{\epsilon})}{d\mu(I_{\epsilon})}$$

with $I_{\epsilon} = [t, t + \epsilon] \cap [0, \tau]$.

We define

$$N' = \{t \in [0, \tau]; \dot{u}(t+0) = \dot{u}(t-0) \neq \dot{u}(t)\}$$

(we may observe that the convention (24) implies that $0 \notin N'$ and $\tau \notin N'$). Since $\dot{u} = v$ belongs to $BV(0, \tau; \mathbb{R}^d)$, $N'$ is, at most, a countable subset of $[0, \tau]$ and is negligible with respect to $|d\dot{u}|$.

Finally, let $N_0 = \{t \in \{0\} \cup \{\tau\}; \dot{u}$ is continuous at $t\}$. The set $N_0$ is finite (it contains at most the two points $t=0$ and $t=\tau$), so it is negligible with respect to $|d\dot{u}|$, and it follows that $N \cup N' \cup N_0$ is also negligible with respect to $d\mu$. We have:

**Proposition 4.** Let $t \in [0, \tau] \setminus (N \cup N' \cup N_0)$ such that $\dot{u}$ is continuous at $t$. Then

$$M(u(t)) v'_{\mu}(t) - g(t, u(t), \dot{u}(t)) t'_{\mu}(t) \in -N_K(u(t)). \quad (25)$$

**Proof.** Let $t \in [0, \tau] \setminus (N \cup N' \cup N_0)$ such that $\dot{u}$ is continuous at $t$. Then $t \in (0, \tau)$; for simplicity we will denote $\bar{u} = u(t)$ in the remainder of the proof. By definition of $N_K(\bar{u})$, (25) is equivalent to

$$(g(t, \bar{u}, \dot{u}(t)) t'_{\mu}(t) - M(\bar{u}) v'_{\mu}(t), w) \leq 0$$

for all $w \in T_K(\bar{u}) = \{w \in \mathbb{R}^d; (\nabla f_{\alpha}(\bar{u}), w) \geq 0, \forall \alpha \in J(\bar{u})\}$.

First, let us observe that there exists $r_{\bar{u}} > 0$ such that

$$J(q) \subset J(\bar{u}) \quad \forall q \in \bar{B}(\bar{u}, r_{\bar{u}}).$$

Indeed, for all $\alpha \in \{1, \dots, v\} \setminus J(\bar{u})$ we have $f_{\alpha}(\bar{u}) > 0$ and, by continuity of the mappings $f_{\alpha}$ ($1 \leq \alpha \leq v$), there exists $r_{\bar{u}} > 0$ such that

$$f_{\alpha}(q) \geq \frac{f_{\alpha}(\bar{u})}{2} \quad \forall q \in \bar{B}(\bar{u}, r_{\bar{u}}), \quad \forall \alpha \in \{1, \dots, v\} \setminus J(\bar{u}).$$
Let us consider $\tilde{T}_K(\bar{u})$ defined by

$$ \tilde{T}_K(\bar{u}) = \begin{cases} \{w \in \mathbb{R}^d; (\nabla f_\alpha(\bar{u}), w) > 0 \ \forall \alpha \in J(\bar{u})\} & \text{if } J(\bar{u}) \neq \emptyset, \\ \mathbb{R}^d & \text{otherwise.} \end{cases} $$

Let $w \in \tilde{T}_K(\bar{u})$. If $J(\bar{u}) \neq \emptyset$, the continuity of the mappings $\nabla f_\alpha$ ($1 \le \alpha \le \nu$) implies that there exists $r_w \in (0, r_{\bar{u}}]$ such that

$$ (\nabla f_\alpha(q), w) \ge 0 \quad \forall \alpha \in J(\bar{u}), \quad \forall q \in \bar{B}(\bar{u}, r_w), $$

and thus $w \in T_K(q)$ for all $q \in \bar{B}(\bar{u}, r_w)$. If $J(\bar{u}) = \emptyset$, we still have $w \in T_K(q)$ for all $q \in \bar{B}(\bar{u}, r_w)$ if we choose $r_w = r_{\bar{u}}$.

Using the continuity of $u$ and the uniform convergence of $(u_{h_i})_{i \in \mathbb{N}}$ to $u$ on $[0, \tau]$, there exists $\tilde{\varepsilon}_w \in (0, \min(t, \frac{\tau-t}{2}))$ such that, for all $\varepsilon \in (0, \tilde{\varepsilon}_w]$, there exists $h_\varepsilon \in (0, h_1^*]$ such that

$$ \begin{gathered} u(s) \in \bar{B}\left(\bar{u}, \frac{r_w}{3}\right) \quad \forall s \in [t, t+\varepsilon], \\ h_\varepsilon \le \min\left(\frac{r_w}{3C_0}, \frac{\varepsilon}{3}\right), \quad \|u - u_{h_i}\|_{C^0([0, \tau]; \mathbb{R}^d)} \le \frac{r_w}{3} \quad \forall h_i \in (0, h_\varepsilon]. \end{gathered} $$

It follows that for all $\varepsilon \in (0, \tilde{\varepsilon}_w]$ and for all $h_i \in (0, h_\varepsilon]$

$$ u_{h_i}(s) \in \bar{B}\left(\bar{u}, \frac{2r_w}{3}\right) \quad \forall s \in [t, t+\varepsilon], $$

and

$$ U^{n+1} \in \bar{B}(\bar{u}, r_w) \quad \forall nh_i \in [t, t+\varepsilon]. $$

Now let $\varepsilon \in (0, \tilde{\varepsilon}_w]$ and $h_i \in (0, h_\varepsilon]$. We define $j$ and $k$ by

$$ j = \left\lfloor \frac{t}{h_i} \right\rfloor, \quad k = \left\lfloor \frac{t+\varepsilon}{h_i} \right\rfloor. $$

We have

$$ 0 < t_j = jh_i \le t < t_{j+1} < \dots < t_k = kh_i \le t + \varepsilon < t_{k+1} < \tau. $$

From Lemma 1 we know that, for all $n \in \{j+1, \dots, k\}$, we have

$$ (V^{n-1} - V^n + h_i F^n, w)_{M(U^n)} \le 0 $$

since $w \in T_K(U^{n+1})$, and by summation

$$ \begin{aligned} & \sum_{n=j+1}^{k} h_i (M(U^n)F^n, w) + \sum_{n=j+1}^{k} \left( M(U^{n-1})V^{n-1} - M(U^n)V^n, w \right) \\ & \qquad + \sum_{n=j+1}^{k} \left( M(U^n)V^{n-1} - M(U^{n-1})V^{n-1}, w \right) \le 0. \end{aligned} \quad (26) $$
The last term can be easily estimated as $O(\epsilon + h_i)$. Indeed, the Lipschitz property of the mapping $M$ on $B_1$ implies that

$$ \| M(U^n) - M(U^{n-1}) \| \le L_M h_i |V^{n-1}| \le h_i L_M C_0. $$

It follows that

$$ \begin{aligned} & \left| \sum_{n=j+1}^{k} (M(U^n)V^{n-1} - M(U^{n-1})V^{n-1}, w) \right| \\ & \le \sum_{n=j+1}^{k} \| M(U^n) - M(U^{n-1}) \| C_0 |w| \\ & \le (k-j)L_M h_i C_0^2 |w| \le (\epsilon + h_i)L_M C_0^2 |w|. \end{aligned} $$

The second term of the left-hand side of (26) is a telescopic sum which can be rewritten as

$$ \begin{aligned} & (M(U^j)V^j - M(U^k)V^k, w) = (M(u_{h_i}(t))(v_{h_i}(t) - v_{h_i}(t+\epsilon)), w) \\ & \qquad + ((M(U^j) - M(u_{h_i}(t)))V^j, w) \\ & \qquad + ((M(u_{h_i}(t)) - M(U^k))V^k, w). \end{aligned} $$

Once again, the last two terms can be estimated by using the Lipschitz property of $M$ on $B_1$:

$$ \left| \left( (M(U^j) - M(u_{h_i}(t)))V^j + (M(u_{h_i}(t)) - M(U^k))V^k, w \right) \right| \le L_M C_0 |w| \left( |u_{h_i}(j h_i) - u_{h_i}(t)| + |u_{h_i}(t) - u_{h_i}(k h_i)| \right) \le L_M C_0^2 |w| (h_i + \epsilon). $$

Moreover, with (21) and (22), we have

$$ \begin{aligned} & \lim_{h_i \to 0} M(u_{h_i}(t))(v_{h_i}(t) - v_{h_i}(t+\epsilon)) = M(u(t))(v(t) - v(t+\epsilon)) \\ & = M(u(t))(\dot{u}(t) - \dot{u}(t+\epsilon)). \end{aligned} $$

Let us prove now that

$$ \lim_{h_i \to 0} \sum_{n=j+1}^{k} h_i (M(U^n)F^n, w) = \int_{t}^{t+\epsilon} (g(s, u(s), \dot{u}(s)), w) \, ds $$

for all $\epsilon \in (0, \tilde{\epsilon}_w]$.

Indeed, let $\epsilon \in (0, \tilde{\epsilon}_w]$. For all $h_i \in (0, h_\epsilon]$ and $n \in \{j+1, \dots, k\}$, we have $F^n = F(nh_i, U^n, V^{n-1}, h_i)$ and with (H5)

$$ F(nh_i, U^n, V^{n-1}, 0) = M^{-1}(U^n)g(nh_i, u_{h_i}(nh_i), v_{h_i}(s)) \quad \forall s \in [(n-1)h_i, nh_i). $$
It follows that

$$
\begin{align}
& \sum_{n=j+1}^{k} h_i (M(U^n)F^n, w) - \sum_{n=j+1}^{k} \int_{(n-1)h_i}^{nh_i} (g(nh_i, u_{h_i}(nh_i), v_{h_i}(s)), w) \, ds \nonumber \\
&= \sum_{n=j+1}^{k} h_i \left( F(nh_i, U^n, V^{n-1}, h_i) - F(nh_i, U^n, V^{n-1}, 0), w \right)_{M(U^n)} \tag{27}
\end{align}
$$

In order to estimate the right-hand side of (27), we denote by $\omega_F$ the modulus of continuity of $F$ on the compact set $[0, T] \times B_1 \times \bar{B}(0, C_0) \times [0, h^*]$ and we get

$$
\left| \sum_{n=j+1}^{k} h_i \left( F(nh_i, U^n, V^{n-1}, h_i) - F(nh_i, U^n, V^{n-1}, 0), w \right)_{M(U^n)} \right| \\
\le \sum_{n=j+1}^{k} h_i \| M(U^n) \|_{\omega_F(h_i)} |w| \le (\epsilon + h_i) \lambda_{max, B_1} \omega_F(h_i) |w|. \quad (28)
$$

Furthermore

$$
\begin{align}
& \sum_{n=j+1}^{k} \int_{(n-1)h_i}^{nh_i} (g(nh_i, u_{h_i}(nh_i), v_{h_i}(s)), w) \, ds \nonumber \\
& \quad - \int_{t}^{t+\epsilon} (g(s, u_{h_i}(s), v_{h_i}(s)), w) \, ds \nonumber \\
& = \sum_{n=j+1}^{k} \int_{(n-1)h_i}^{nh_i} (g(nh_i, u_{h_i}(nh_i), v_{h_i}(s)) - g(s, u_{h_i}(s), v_{h_i}(s)), w) \, ds \nonumber \\
& \quad - \int_{kh_i}^{t+\epsilon} (g(s, u_{h_i}(s), v_{h_i}(s)), w) \, ds + \int_{jh_i}^{t} (g(s, u_{h_i}(s), v_{h_i}(s)), w) \, ds. \tag{29}
\end{align}
$$

Recalling that $(u_{h_i}(s), v_{h_i}(s)) \in B_1 \times \bar{B}(0, C_0)$ for all $s \in [0, \tau]$ and for all $h_i \in (0, h_1^*)$, we obtain the following estimates for the second and third terms of the right-hand side of (29):

$$
\begin{gather*}
\left| \int_{kh_i}^{t+\epsilon} (g(s, u_{h_i}(s), v_{h_i}(s)), w) \, ds \right| \le |t + \epsilon - kh_i| C_g |w| \le h_i C_g |w|, \\
\left| \int_{jh_i}^{t} (g(s, u_{h_i}(s), v_{h_i}(s)), w) \, ds \right| \le |t - jh_i| C_g |w| \le h_i C_g |w|,
\end{gather*}
$$

with $C_g = \sup\{|g(s, q, v)|; (s, q, v) \in [0, T] \times B_1 \times \bar{B}(0, C_0)\}$.

In order to estimate the first term of the right-hand side of (29), we introduce
$\omega_g$ the modulus of continuity of $g$ on $[0, T] \times B_1 \times \bar{B}(0, C_0)$. Observing that

$$
|u_{h_i}(nh_i) - u_{h_i}(s)| \le C_0|nh_i - s| \le C_0h_i
$$
for all $s \in [(n-1)h_i, nh_i)$ and for all $n \in \{j+1, \dots, k\}$, we get

$$|g(nh_i, u_{h_i}(nh_i), v_{h_i}(s)) - g(s, u_{h_i}(s), v_{h_i}(s))| \le \omega_g(C_0h_i) + \omega_g(h_i)$$

for all $s \in [(n-1)h_i, nh_i)$ and for all $n \in \{j+1, \dots, k\}$.

Hence

$$\left| \sum_{n=j+1}^{k} \int_{(n-1)h_i}^{nh_i} \left( g(nh_i, u_{h_i}(nh_i), v_{h_i}(s)), w \right) \mathrm{d}s - \int_{t}^{t+\varepsilon} \left( g(s, u_{h_i}(s), v_{h_i}(s)), w \right) \mathrm{d}s \right| \\ \le 2h_i C_g |w| + (\omega_g(C_0h_i) + \omega_g(h_i))(k-j)h_i |w| \\ \le 2h_i C_g |w| + (\omega_g(C_0h_i) + \omega_g(h_i))(\varepsilon + h_i) |w|. \quad (30)$$

Then recalling that

$$u_{h_i}(s) \to_{h_i \to 0} u(s) \quad \text{for all } s \in [0, \tau],$$

and

$$v_{h_i}(s) \to_{h_i \to 0} v(s) = \dot{u}(s) \quad \text{for a.a. } s \in [0, \tau],$$

we infer from Lebesgue's theorem that

$$\lim_{h_i \to 0} \int_t^{t+\varepsilon} \left( g(s, u_{h_i}(s), v_{h_i}(s)), w \right) \mathrm{d}s = \int_t^{t+\varepsilon} \left( g(s, u(s), \dot{u}(s)), w \right) \mathrm{d}s. \quad (31)$$

Finally, combining (31), (30) and (29), we obtain

$$\lim_{h_i \to 0} \sum_{n=j+1}^{k} \int_{(n-1)h_i}^{nh_i} \left( g(nh_i, u_{h_i}(nh_i), v_{h_i}(s)), w \right) \mathrm{d}s \\ = \int_{t}^{t+\varepsilon} \left( g(s, u(s), \dot{u}(s)), w \right) \mathrm{d}s$$

and with (27) and (28) we may conclude that

$$\lim_{h_i \to 0} \sum_{n=j+1}^{k} h_i (M(U^n)F^n, w) = \int_{t}^{t+\varepsilon} (g(s, u(s), \dot{u}(s)), w) \mathrm{d}s.$$

Then, passing to the limit as $h_i$ tends to zero in (26), we get

$$\int_{t}^{t+\varepsilon} \left( g(s, u(s), \dot{u}(s)), w \right) \mathrm{d}s + \left( M(u(t))\dot{u}(t) - \dot{u}(t+\varepsilon), w \right) \\ \le 2L_M \varepsilon C_0^2 |w| \quad (32)$$
for all $\varepsilon \in (0, \tilde{\varepsilon}_w]$. Since $\dot{u}$ is continuous at $t$, we have $\dot{u}(t) = \dot{u}(t-0) = \dot{u}(t+0)$. Moreover, since $\dot{u} = v$ and $v$ is continuous, except perhaps on a countable subset of $[0, \tau]$, we may choose a sequence $(\varepsilon_i)_{i \in \mathbb{N}}$ decreasing to zero such that

$$ \varepsilon_i \in (0, \tilde{\varepsilon}_w], \quad v(t + \varepsilon_i) = \dot{u}(t + \varepsilon_i) = \dot{u}(t + \varepsilon_i + 0) \quad \forall i \in \mathbb{N}. $$

It follows that

$$ \dot{u}(t) - \dot{u}(t + \varepsilon_i) = \dot{u}(t - 0) - \dot{u}(t + \varepsilon_i + 0) = -d\dot{u}([t, t + \varepsilon_i]) \quad \forall i \in \mathbb{N}. $$

Multiplying (32) by $\frac{1}{d\mu([t, t+\varepsilon_i])}$ and passing to the limit as $\varepsilon_i$ tends to zero, we obtain

$$ (g(t, u(t), \dot{u}(t)), w) t'_{\mu}(t) - (M(u(t)) v'_{\mu}(t), w) \le 0 \quad \forall w \in \tilde{T}_K(\bar{u}). $$

Finally, observing that $\tilde{T}_K(\bar{u})$ is dense in $T_K(\bar{u})$ we may conclude. $\square$

Let us now consider $t \in [0, \tau] \setminus (N \cup N' \cup N_0)$ such that $\dot{u}$ is discontinuous at $t$. Then $\dot{u}(t-0) \ne \dot{u}(t+0)$ and $d\dot{u}$ possesses a Dirac mass at $t$. Thus $\{t\}$ is not negligible anymore with respect to $d\mu$ and (3) is equivalent to

$$ M(u(t)) (\dot{u}(t+0) - \dot{u}(t-0)) \in -N_K(u(t)). $$

This property is a direct consequence of the following proposition:

**Proposition 5.** For all $t \in [0, \tau]$ we have

$$ M(u(t)) (\dot{u}(t+0) - \dot{u}(t-0)) \in -N_K(u(t)). $$

**Proof.** Let $t \in [0, \tau]$ and denote, for simplicity, $\bar{u} = u(t)$. Thanks to the density of $\tilde{T}_K(\bar{u})$ in $T_K(\bar{u})$, we only need to prove that

$$ (M(\bar{u}) (\dot{u}(t-0) - \dot{u}(t+0)), w) \le 0 \quad \forall w \in \tilde{T}_K(\bar{u}). $$

Let $w \in \tilde{T}_K(\bar{u})$. As in the proof of the previous proposition, we define $r_w > 0$ such that

$$ J(q) \subset J(\bar{u}) \quad \text{and} \quad w \in T_K(q) \quad \text{for all } q \in \bar{B}(\bar{u}, r_w). $$

We also define $\tilde{\varepsilon}_w \in (0, \tau/2)$ such that for all $\varepsilon \in (0, \tilde{\varepsilon}_w]$ we have

$$ u(s) \in \bar{B}(\bar{u}, \frac{r_w}{3}) \quad \forall s \in [t - \varepsilon, t + \varepsilon] \cap [0, \tau], $$

and there exists $h_\varepsilon \in (0, \min(h_1^*, r_w/(3C_0), \varepsilon/3)]$ such that

$$ u_{h_i}(s) \in \bar{B}(\bar{u}, \frac{2r_w}{3}) \quad \forall s \in [t - \varepsilon, t + \varepsilon] \cap [0, \tau], \quad \forall h_i \in (0, h_\varepsilon], $$

and

$$ U^{n+1} \in \bar{B}(\bar{u}, r_w) \quad \forall nh_i \in [t - \varepsilon, t + \varepsilon] \cap [0, \tau], \quad \forall h_i \in (0, h_\varepsilon]. $$
Let $\epsilon \in (0, \tilde{\epsilon}_w]$ and $h_i \in (0, h_\epsilon]$. We define $t_\epsilon^- = \max(t-\epsilon, 0)$, $t_\epsilon^+ = \min(t+\epsilon, \tau)$ and

$$j = \left\lfloor \frac{t_{\epsilon}^{-}}{h_{i}} \right\rfloor, \quad k = \left\lfloor \frac{t_{\epsilon}^{+}}{h_{i}} \right\rfloor$$

that is, we have

$$0 \le t_j = jh_i \le t_\epsilon^- < t_{j+1} < \dots < t_k = kh_i \le t_\epsilon^+ \le \tau.$$

It follows that

$$v_{h_i}(t_{\epsilon}^{-}) = V^{j}, \quad v_{h_i}(t_{\epsilon}^{+}) = V^{k}.$$

We have

$$\begin{aligned} & (M(u_{h_i}(t_\epsilon^-)) v_{h_i}(t_\epsilon^-) - M(u_{h_i}(t_\epsilon^+)) v_{h_i}(t_\epsilon^+), w) \\ & = (M(U^j)V^j - M(U^k)V^k, w) \\ & + ((M(u_{h_i}(t_\epsilon^-)) - M(U^j))V^j, w) + ((M(U^k) - M(u_{h_i}(t_\epsilon^+)))V^k, w). \end{aligned} \quad (33)$$

Following the same ideas as in the previous proof, we use the Lipschitz continuity of $M$ on $B_1$ to estimate the last two terms of (33). More precisely,

$$\left| \left( \left( M(u_{h_i}(t_\epsilon^-)) - M(U^j) \right) V^j, w \right) \right| \le |w| |V^j| L_M \left| u_{h_i}(t_\epsilon^-) - U^j \right| \le L_M |w| C_0^2 h_i,$$

and with similar computations

$$\left| \left( (M(U^k) - M(u_{h_i}(t_\epsilon^+))) V^k, w \right) \right| \le L_M |w| C_0^2 h_i.$$

We rewrite the first term of (33) as

$$\begin{aligned} (M(U^j)V^j - M(U^k)V^k, w) = & \sum_{n=j+1}^{k} M(U^n)(V^{n-1} - V^n, w) \\ & + \sum_{n=j+1}^{k} \left( (M(U^{n-1}) - M(U^n)) V^{n-1}, w \right), \end{aligned}$$

and, observing that $w \in T_K(U^{n+1})$ for all $n \in \{j+1, \dots, k\}$, we infer from Lemma 1 that

$$\begin{aligned} \sum_{n=j+1}^{k} M(U^n)(V^{n-1} - V^n, w) &\le \sum_{n=j+1}^{k} -h_i (M(U^n)F^n, w) \\ &\le (k-j)h_i \lambda_{max,B_1} C_F |w| \\ &\le (2\epsilon + h_i) \lambda_{max,B_1} C_F |w|. \end{aligned}$$
It follows that

$$ (M(U^j)V^j - M(U^k)V^k, w) \le \sum_{n=j+1}^k \left| \begin{aligned} & \left| \left( (M(U^{n-1}) - M(U^n))V^{n-1}, w \right) \right| \\ & + (2\epsilon + h_i)\lambda_{\max, B_1} C_F |w| \end{aligned} \right| $$

and, using once again the Lipschitz property of $M$, we get

$$ (M(U^j)V^j - M(U^k)V^k, w) \le (2\epsilon + h_i)|w| (\lambda_{\max, B_1} C_F + L_M C_0^2). $$

Finally, we obtain

$$ \begin{aligned} & (M(u_{h_i}(t_{\bar{\epsilon}}^{-})) v_{h_i}(t_{\bar{\epsilon}}^{-}) - M(u_{h_i}(t_{\bar{\epsilon}}^{+})) v_{h_i}(t_{\bar{\epsilon}}^{+}), w) \\ & \le 2L_M |w| C_0^2 h_i + (2\epsilon + h_i) |w| (\lambda_{\max, B_1} C_F + L_M C_0^2), \end{aligned} $$

for all $h_i \in (0, h_\epsilon]$ and for all $\epsilon \in (0, \tilde{\epsilon}_w]$.

Now, passing to the limit as $h_i$ tends to zero, then as $\epsilon$ tends to zero, we may conclude. $\square$

### 3.2. Transmission of the velocity at impacts

With the previous proposition, we observe that $\dot{u}(t+0) = \dot{u}(t-0)$ for all $t \in [0, \tau]$ such that $J(u(t)) = \emptyset$. That is, $\dot{u}$ is continuous at $t$ if $u(t) \in \text{Int}(K)$ and, in this case, the impact law (4) is satisfied. Thus it remains only to prove that

$$ \dot{u}(\bar{t} + 0) = \text{Proj}_{M(u(\bar{t}))} (T_K(u(\bar{t})), \dot{u}(\bar{t} - 0)) \quad (34) $$

for all $\bar{t} \in (0, \tau)$ such that $J(u(\bar{t})) \neq \emptyset$.

In order to also obtain some information on $\dot{u}(0+0)$, we now consider $\bar{t} \in [0, \tau)$ such that $J(u(\bar{t})) \neq \emptyset$. For simplicity, we denote $\bar{u} = u(\bar{t})$ and $\dot{u}^+ = \dot{u}(\bar{t} + 0)$, $\dot{u}^- = \dot{u}(\bar{t} - 0)$. With Proposition 5 we already know that $M(\bar{u})(\dot{u}^- - \dot{u}^+) \in N_K(\bar{u})$, that is, there exist non-positive real numbers $(\mu_\alpha)_{\alpha \in J(\bar{u})}$ such that

$$ M^{1/2}(\bar{u}) (\dot{u}^- - \dot{u}^+) = \sum_{\alpha \in J(\bar{u})} \mu_\alpha e_\alpha(\bar{u}), $$

where we recall that

$$ e_\alpha(\bar{u}) = \frac{M^{-1/2}(\bar{u}) \nabla f_\alpha(\bar{u})}{|M^{-1/2}(\bar{u}) \nabla f_\alpha(\bar{u})|} \quad \forall \alpha \in J(\bar{u}). $$

Moreover, since $u(t) \in K$ for all $t \in [0, \tau]$, we have $\dot{u}^+ \in T_K(\bar{u})$ and (34) reduces to

$$ (\dot{u}^- - \dot{u}^+, \dot{u}^+)_{M(\bar{u})} = 0, $$

that is,

$$ (e_\alpha(\bar{u}), M^{1/2}(\bar{u})\dot{u}^+) = 0 \quad \text{for all } \alpha \in J(\bar{u}) \text{ such that } \mu_\alpha \neq 0. $$
From assumption (H3) we know that $(\nabla f_{\alpha}(\bar{u}))_{\alpha \in J(\bar{u})}$ is linearly independent. It follows that $(e_{\alpha}(\bar{u}))_{\alpha \in J(\bar{u})}$ is also linearly independent and there exist $(e_{\beta})_{\beta \in \{1, \dots, d\} \setminus J(\bar{u})}$ such that $\{e_{\alpha}(\bar{u}); \alpha \in J(\bar{u})\} \cup \{e_{\beta}; \beta \in \{1, \dots, d\} \setminus J(\bar{u})\}$ is a basis of $\mathbb{R}^d$ and $|e_{\beta}| = 1$ for all $\beta \in \{1, \dots, d\} \setminus J(\bar{u})$.

Using Lemma 7 (see Appendix) we know that there exists $r_{\bar{u}} \in (0, r_{B_1}]$ such that

$$J(q) \subset J(\bar{u}) \quad \forall q \in \bar{B}(\bar{u}, r_{\bar{u}}) \qquad (35)$$

and

$$|M^{-1/2}(q) \nabla f_{\alpha}(q)| \geq m_{B_1} > 0 \quad \forall q \in \bar{B}(\bar{u}, r_{\bar{u}}), \forall \alpha \in J(\bar{u}).$$

Thus, for all $\alpha \in \{1, \dots, d\}$ and for all $q \in \bar{B}(\bar{u}, r_{\bar{u}})$ we define

$$v_{\alpha}(q) = \begin{cases} \frac{M^{-1/2}(q) \nabla f_{\alpha}(q)}{|M^{-1/2}(q) \nabla f_{\alpha}(q)|} & \text{if } \alpha \in J(\bar{u}), \\ e_{\alpha} & \text{if } \alpha \notin J(\bar{u}). \end{cases}$$

From (35) we infer that $v_{\alpha}(q) = e_{\alpha}(q)$ for all $\alpha \in J(q)$, for all $q \in \bar{B}(\bar{u}, r_{\bar{u}}) \cap K$. Moreover, the continuity of $M^{-1/2}$ and $\nabla f_{\alpha}$ ($1 \leq \alpha \leq \nu$) implies that, possibly decreasing $r_{\bar{u}}$, $(v_{\alpha}(q))_{1 \leq \alpha \leq d}$ is a basis of $\mathbb{R}^d$ for all $q \in \bar{B}(\bar{u}, r_{\bar{u}})$. We define the dual basis $(w_{\alpha}(q))_{1 \leq \alpha \leq d}$ for all $q \in \bar{B}(\bar{u}, r_{\bar{u}})$. From Lemma 7, we know that the vectors $(w_{\alpha}(q))_{1 \leq \alpha \leq d}$ are bounded independently of $q$ by a constant $C_{*,\bar{u}}$ and, since the mappings $M^{-1/2}$ and $\nabla f_{\alpha}$ ($1 \leq \alpha \leq \nu$) are locally Lipschitz continuous, the mappings $v_{\alpha}$ and $w_{\alpha}$ ($1 \leq \alpha \leq d$) are also Lipschitz continuous on $\bar{B}(\bar{u}, r_{\bar{u}})$; we let $L_{\bar{u}} \in \mathbb{R}_{*}^+$ be such that, for all $\alpha \in \{1, \dots, d\}$ and for all $(q, q') \in \bar{B}(\bar{u}, r_{\bar{u}})^2$

$$|v_{\alpha}(q) - v_{\alpha}(q')| \leq L_{\bar{u}}|q - q'|, \quad |w_{\alpha}(q) - w_{\alpha}(q')| \leq L_{\bar{u}}|q - q'|.$$

Finally, from the continuity of $u$ and the uniform convergence of $(u_{h_i})_{i \in \mathbb{N}}$ to $u$ on $[0, \tau]$, we infer that there exist $\bar{\epsilon} \in (0, \frac{\tau - \bar{\tau}}{2}]$ and $h_2^* \in (0, \min(h_1^*, \frac{\bar{\epsilon}}{3}, \frac{r_{\bar{u}}}{3C_0})]$ such that

$$u(t) \in \bar{B} \left( \bar{u}, \frac{r_{\bar{u}}}{3} \right) \quad \forall t \in [\bar{t} - \bar{\epsilon}, \bar{t} + \bar{\epsilon}] \cap [0, \tau], \\ \|u - u_{h_i}\|_{C^0([0, \tau]; \mathbb{R}^d)} \leq \frac{r_{\bar{u}}}{3} \quad \forall h_i \in (0, h_2^*].$$

It follows that

$$U^{n+1}, U^n \in \bar{B}(\bar{u}, r_{\bar{u}}) \quad \forall nh_i \in [t - \bar{\epsilon}, t + \bar{\epsilon}] \cap [0, \tau], \forall h_i \in (0, h_2^*]. \qquad (36)$$

We begin with the following lemma.

**Lemma 3.** Let $\alpha \in J(\bar{u})$ such that $\mu_{\alpha} \neq 0$. Then, for all $\varepsilon_1 \in (0, \bar{\epsilon}]$ there exists $h_{\varepsilon_1} \in (0, \min(h_2^*, \varepsilon_1/3)]$ such that for all $h_i \in (0, h_{\varepsilon_1}]$, there exists $nh_i \in [\bar{t} - \varepsilon_1, \bar{t} + \varepsilon_1] \cap [0, \tau]$ such that $f_{\alpha}(U^{n+1}) \leq 0$.
**Proof.** Let us assume that the announced result does not hold, that is, assume that there exists $\varepsilon_1 \in (0, \bar{\varepsilon}]$ such that, for all $h_{\varepsilon_1} \in (0, \min(h_2^*, \varepsilon_1/3)]$ there exists $h_i \in (0, h_{\varepsilon_1}]$ such that $f_\alpha(U^{n+1}) > 0$ for all $nh_i \in [\bar{t} - \varepsilon_1, \bar{t} + \varepsilon_1] \cap [0, \tau]$.

Hence, we can extract from $(h_i)_{i \in \mathbb{N}}$ a subsequence denoted $(h_{\varphi(i)})_{i \in \mathbb{N}}$ such that $h_{\varphi(i)} \in (0, \min(h_2^*, \varepsilon_1/3)]$, $(h_{\varphi(i)})_{i \in \mathbb{N}}$ decreases to zero and

$$f_{\alpha}(U^{n+1}) > 0 \quad \forall nh_{\varphi(i)} \in [\bar{t} - \varepsilon_1, \bar{t} + \varepsilon_1] \cap [0, \tau] \quad (37)$$

for all $i \in \mathbb{N}$.

For all $\varepsilon \in (0, \varepsilon_1]$, let us establish the following estimate:

$$\left| \left( M^{1/2} \left( u_{h_{\varphi(i)}}(t_{\varepsilon}^{-}) \right) v_{h_{\varphi(i)}}(t_{\varepsilon}^{-}) - M^{1/2} \left( u_{h_{\varphi(i)}}(t_{\varepsilon}^{+}) \right) v_{h_{\varphi(i)}}(t_{\varepsilon}^{+}), w_{\alpha}(\bar{u}) \right) \right| \\ \le \mathcal{O} \left( \varepsilon + h_{\varphi(i)} + \|u - u_{h_{\varphi(i)}}\|_{C^0([0, \tau]; \mathbb{R}^d)} \right)$$

where $t_{\varepsilon}^{-} = \max(\bar{t} - \varepsilon, 0)$ and $t_{\varepsilon}^{+} = \min(\bar{t} + \varepsilon, \tau)$. Then, by passing to the limit when $i$ tends to $+\infty$, we will infer with (21) and (22) that

$$\left| \left( M^{1/2} \left( u(t_{\varepsilon}^{-}) \right) v(t_{\varepsilon}^{-}) - M^{1/2} \left( u(t_{\varepsilon}^{+}) \right) v(t_{\varepsilon}^{+}), w_{\alpha}(\bar{u}) \right) \right| \le \mathcal{O}(\varepsilon)$$

and, when $\varepsilon$ tends to zero, we will obtain

$$\left| \left( M^{1/2}(\bar{u}) \left( \dot{u}(\bar{t} - 0) - \dot{u}(\bar{t} + 0) \right), w_{\alpha}(\bar{u}) \right) \right| = |\mu_{\alpha}| \le 0$$

which gives a contradiction.

Let $\varepsilon \in (0, \varepsilon_1]$. There exists $i_\varepsilon \in \mathbb{N}$ such that, for all $i \ge i_\varepsilon$ we have $h_{\varphi(i)} \in (0, \varepsilon/2)$ and we define

$$n_i = \left\lfloor \frac{t_{\varepsilon}^{-}}{h_{\varphi(i)}} \right\rfloor, \quad p_i = \left\lfloor \frac{t_{\varepsilon}^{+}}{h_{\varphi(i)}} \right\rfloor.$$

Then, $n_i + 1 < p_i$ and for all $n \in \{n_i + 1, \dots, p_i\}$ we have $nh_{\varphi(i)} \in [t_{\varepsilon}^{-}, t_{\varepsilon}^{+}]$. We infer from Lemma 1 that, for all $n \in \{n_i + 1, \dots, p_i\}$

$$M(U^n) \left( V^{n-1} - V^n + h_{\varphi(i)} F^n \right) \in N_K(U^{n+1}).$$

If $J(U^{n+1}) \ne \emptyset$, there exist non-positive real numbers $(\mu_\beta^n)_{\beta \in J(U^{n+1})}$ such that

$$M(U^n) \left( V^{n-1} - V^n + h_{\varphi(i)} F^n \right) = \sum_{\beta \in J(U^{n+1})} \mu_{\beta}^n M^{1/2}(U^{n+1}) e_{\beta}(U^{n+1}).$$

From (36), we obtain $e_\beta(U^{n+1}) = v_\beta(U^{n+1})$ for all $\beta \in J(U^{n+1})$ and

$$\begin{aligned} |\mu_{\beta}^{n}| &= \left| \left( M(U^{n}) (V^{n-1} - V^{n} + h_{\varphi(i)} F^{n}), M^{-1/2}(U^{n+1}) w_{\beta}(U^{n+1}) \right) \right| \\ &\le \frac{\lambda_{\max, B_1}}{\sqrt{\lambda_{\min, B_1}}} (2C_0 + h_2^* C_F) C_{*, \bar{u}} \le 3 \frac{\lambda_{\max, B_1}}{\sqrt{\lambda_{\min, B_1}}} C_0 C_{*, \bar{u}} \quad \forall \beta \in J(U^{n+1}). \end{aligned}$$
From now on, let us denote

$$C_2' = 3 \frac{\lambda_{\max, B_1}}{\sqrt{\lambda_{\min, B_1}}} C_0 C_{*, \bar{u}}.$$

With (37) we know that $\alpha \notin J(U^{n+1})$, thus

$$ \begin{aligned} & |(M^{1/2}(U^{n+1})(V^{n-1} - V^n + h_{\varphi(i)}F^n), w_\alpha(U^{n+1}))| \\ &= \left| \sum_{\beta \in J(U^{n+1})} \mu_\beta^n (M^{1/2}(U^{n+1})M^{-1}(U^n)M^{1/2}(U^{n+1})v_\beta(U^{n+1}), w_\alpha(U^{n+1})) \right| \\ &\le \sum_{\beta \in J(U^{n+1})} |\mu_\beta^n| \|M^{1/2}(U^{n+1})\|^2 \|M^{-1}(U^{n+1}) - M^{-1}(U^n)\| |v_\beta(U^{n+1})| |w_\alpha(U^{n+1})| \\ &\le \nu C_2' C_{*, \bar{u}} \lambda_{\max, B_1} L_{M-1} |U^{n+1} - U^n| \le \nu C_2' C_0 C_{*, \bar{u}} \lambda_{\max, B_1} L_{M-1} h_{\varphi(i)} \end{aligned} $$

for all $n \in \{n_i + 1, \dots, p_i\}$, if $J(U^{n+1}) \neq \emptyset$.

If $J(U^{n+1}) = \emptyset$, this last inequality remains true since $V^{n-1} - V^n + h_{\varphi(i)}F^n = 0$ if $J(U^{n+1}) = \emptyset$.

It follows that, for all $n \in \{n_i + 1, \dots, p_i\}$

$$ \begin{aligned} & (M^{1/2}(U^{n_i+1})V^{n_i} - M^{1/2}(U^{p_i+1})V^{p_i}, w_\alpha(\bar{u})) \\ &= \sum_{n=n_i+1}^{p_i} (M^{1/2}(U^n)V^{n-1} - M^{1/2}(U^{n+1})V^n, w_\alpha(\bar{u})) \\ &= \sum_{n=n_i+1}^{p_i} (M^{1/2}(U^{n+1})(V^{n-1} - V^n + h_{\varphi(i)}F^n), w_\alpha(U^{n+1})) \\ &\quad + \sum_{n=n_i+1}^{p_i} ((M^{1/2}(U^n) - M^{1/2}(U^{n+1}))V^{n-1}, w_\alpha(\bar{u})) \\ &\quad - \sum_{n=n_i+1}^{p_i} h_{\varphi(i)} (M^{1/2}(U^{n+1})F^n, w_\alpha(U^{n+1})) \\ &\quad + \sum_{n=n_i+1}^{p_i} (M^{1/2}(U^{n+1})(V^{n-1} - V^n), w_\alpha(\bar{u}) - w_\alpha(U^{n+1})) \end{aligned} $$

which yields

$$ \begin{aligned} & \left| (M^{1/2}(U^{n_i+1})V^{n_i} - M^{1/2}(U^{p_i+1})V^{p_i}, w_\alpha(\bar{u})) \right| \\ &\le \sum_{n=n_i+1}^{p_i} \nu C_2' C_0 C_{*, \bar{u}} \lambda_{\max, B_1} L_{M-1} h_{\varphi(i)} \end{aligned} $$
$$
\begin{align*}
&+ \sum_{n=n_i+1}^{p_i} C_0 C_{*,\bar{u}} L_{M^{1/2}} h_{\varphi(i)} |V^n| + \sum_{n=n_i+1}^{p_i} h_{\varphi(i)} C_{*,\bar{u}} C_F \sqrt{\lambda_{\max,B_1}} \\
&+ \sum_{n=n_i+1}^{p_i} \sqrt{\lambda_{\max,B_1}} |V^n - V^{n-1}| |w_{\alpha}(\bar{u}) - w_{\alpha}(U^{n+1})|.
\end{align*}
$$

Then, observing that for all $n \in \{n_i + 1, \dots, p_i\}$

$$
\begin{align*}
|\bar{u} - U^{n+1}| &\le |u(\bar{t}) - u_{h_{\varphi(i)}}(\bar{t})| + |u_{h_{\varphi(i)}}(\bar{t}) - u_{h_{\varphi(i)}}(nh_{\varphi(i)})| + h_{\varphi(i)}|V^n| \\
&\le \|u - u_{h_{\varphi(i)}}\|_{C^0([0, \tau]; \mathbb{R}^d)} + C_0(\varepsilon + h_{\varphi(i)}),
\end{align*}
$$

we can estimate $|w_\alpha(\bar{u}) - w_\alpha(U^{n+1})|$ as

$$
L_{\bar{u}} \left( \|u - u_{h_{\varphi(i)}}\|_{C^0([0, \tau]; \mathbb{R}^d)} + C_0(\varepsilon + h_{\varphi(i)}) \right).
$$

Hence, with the estimate of the discrete accelerations obtained at Proposition 3,
we get

$$
\begin{align}
& |(M^{1/2}(U^{n_i+1})V^{n_i} - M^{1/2}(U^{p_i+1})V^{p_i}, w_\alpha(\bar{u}))| \notag \\
&\le (p_i - n_i)h_{\varphi(i)} C_{*,\bar{u}} \left( \nu C_2'C_0 \lambda_{\max,B_1} L_{M^{-1}} + C_0^2 L_{M^{1/2}} + C_F \sqrt{\lambda_{\max,B_1}} \right) \notag \\
&\quad + \sqrt{\lambda_{\max,B_1}} L_{\bar{u}} \left( \|u - u_{h_{\varphi(i)}}\|_{C^0([0,\tau];\mathbb{R}^d)} + C_0(\varepsilon + h_{\varphi(i)}) \right) \sum_{n=n_i+1}^{p_i} |V^n - V^{n-1}| \notag \\
&= \mathcal{O} \left( \varepsilon + h_{\varphi(i)} + \|u - u_{h_{\varphi(i)}}\|_{C^0([0,\tau];\mathbb{R}^d)} \right). \tag{38}
\end{align}
$$

But $V^{n_i} = v_{h_{\varphi(i)}}(t_{\varepsilon}^{-})$, $V^{p_i} = v_{h_{\varphi(i)}}(t_{\varepsilon}^{+})$ and

$$
\begin{align}
& \left| \left( M^{1/2} (u_{h_{\varphi(i)}}(t_{\varepsilon}^{-})) v_{h_{\varphi(i)}}(t_{\varepsilon}^{-}) - M^{1/2} (u_{h_{\varphi(i)}}(t_{\varepsilon}^{+})) v_{h_{\varphi(i)}}(t_{\varepsilon}^{+}), w_{\alpha}(\bar{u}) \right) \right. \nonumber \\
& \qquad \left. - \left( M^{1/2} (U^{n_i+1}) V^{n_i} - M^{1/2} (U^{p_i+1}) V^{p_i}, w_{\alpha}(\bar{u}) \right) \right| \nonumber \\
& \le \left( \begin{aligned}[t]
& \left| M^{1/2} (u_{h_{\varphi(i)}}(t_{\varepsilon}^{-})) - M^{1/2} (U^{n_i+1}) \right| \\
& + \left| M^{1/2} (u_{h_{\varphi(i)}}(t_{\varepsilon}^{+})) - M^{1/2} (U^{p_i+1}) \right| \end{aligned} \right) C_0 C_{*,\bar{u}} \nonumber \\
& \le 2 L_{M^{1/2}} C_0^2 C_{*,\bar{u}} h_{\varphi(i)}. \tag{39}
\end{align}
$$

Finally, from (38), (39) we obtain

$$
\begin{align*}
& \left| \left( M^{1/2} (u_{h_{\varphi(i)}}(t_{\varepsilon}^{-})) v_{h_{\varphi(i)}}(t_{\varepsilon}^{-}) - M^{1/2} (u_{h_{\varphi(i)}}(t_{\varepsilon}^{+})) v_{h_{\varphi(i)}}(t_{\varepsilon}^{+}), w_{\alpha}(\bar{u}) \right) \right| \\
&= \mathcal{O} \left( \varepsilon + h_{\varphi(i)} + \|u - u_{h_{\varphi(i)}}\|_{C^0([0, \tau]; \mathbb{R}^d)} \right)
\end{align*}
$$

for all $i \ge i_\varepsilon$ and for all $\varepsilon \in (0, \varepsilon_1]$, which enables us to conclude. $\square$
Let us now prove that

$$ (e_{\alpha}(\bar{u}), M^{1/2}(\bar{u})\dot{u}^{+}) = 0 \quad \text{for all } \alpha \in J(\bar{u}) \text{ such that } \mu_{\alpha} \neq 0. $$

**Lemma 4.** Let $\alpha \in J(\bar{u})$ be such that $\mu_\alpha \neq 0$. Then

$$ (e_{\alpha}(\bar{u}), M^{1/2}(\bar{u})\dot{u}^{+}) = 0. $$

**Proof.** Let $\alpha \in J(\bar{u})$ such that $\mu_\alpha \neq 0$. Since $\dot{u}^+ \in T_K(\bar{u})$ we have $(M^{1/2}(\bar{u})\dot{u}^+, e_\alpha(\bar{u})) = (M^{1/2}(\bar{u})\dot{u}^+, v_\alpha(\bar{u})) \ge 0$ and it remains to prove that $(M^{1/2}(\bar{u})\dot{u}^+, v_\alpha(\bar{u})) \le 0$. The main idea of the proof is to obtain an estimate of $(M^{1/2}(u(\bar{t} + \varepsilon))v(\bar{t} + \varepsilon), v_\alpha(u(\bar{t} + \varepsilon)))$ and to pass to the limit when $\varepsilon$ tends to zero.

More precisely, let $\varepsilon \in (0, \bar{\varepsilon}]$. Then

$$ v(\bar{t} + \varepsilon) = \lim_{h_i \to 0} v_{h_i}(\bar{t} + \varepsilon) = \lim_{h_i \to 0} V^{p_i} $$

with $p_i = \lfloor \frac{\bar{t}+\varepsilon}{h_i} \rfloor$ for all $i \in \mathbb{N}$. Observing that

$$ \begin{aligned} |u(\bar{t} + \varepsilon) - U^{p_i+1}| &\le |u(\bar{t} + \varepsilon) - u_{h_i}(\bar{t} + \varepsilon)| + |u_{h_i}(\bar{t} + \varepsilon) - u_{h_i}((p_i+1)h_i)| \\ &\le \|u - u_{h_i}\|_{C^0([0, \tau]; \mathbb{R}^d)} + C_0h_i \end{aligned} $$

the continuity of $v_\alpha$ and $M^{1/2}$ on $\bar{B}(\bar{u}, r_{\bar{u}})$ implies that

$$ \begin{aligned} & (M^{1/2}(u(\bar{t} + \varepsilon))v(\bar{t} + \varepsilon), v_\alpha(u(\bar{t} + \varepsilon))) \\ &= \lim_{h_i \to 0} (M^{1/2}(U^{p_i+1})v_{h_i}(\bar{t} + \varepsilon), v_\alpha(U^{p_i+1})) \\ &= \lim_{h_i \to 0} (M^{1/2}(U^{p_i+1})V^{p_i}, v_\alpha(U^{p_i+1})), \end{aligned} $$

and we will prove that

$$ (M^{1/2}(U^{p_i+1})V^{p_i}, v_\alpha(U^{p_i+1})) \le \mathcal{O} \left( \varepsilon + h_i + \|u - u_{h_i}\|_{C^0([0, \tau]; \mathbb{R}^d)} \right). \quad (40) $$

Let us apply Lemma 3: for all $i$ such that $h_i \in (0, h_\varepsilon]$ we define $N_i$ as the last time step in $[\bar{t} - \varepsilon, \bar{t} + \varepsilon] \cap [0, \tau]$ where the constraint $f_\alpha$ is active, that is,

$$ N_i = \max \left\{ n \in \mathbb{N} ; nh_i \in [\bar{t} - \varepsilon, \bar{t} + \varepsilon] \cap [0, \tau] \text{ and } f_\alpha(U^{n+1}) \le 0 \right\}. $$

Since $\alpha \in J(U^{N_i+1})$, we infer, as in Proposition 1, that

$$ \begin{aligned} & (v_\alpha(U^{N_i+1}), M^{1/2}(U^{N_i+1})V^{N_i}) = (e_\alpha(U^{N_i+1}), M^{1/2}(U^{N_i+1})V^{N_i}) \\ & \le \frac{L_f h_i}{2m_{B_1}} |V^{N_i}|^2 \le \frac{L_f C_0^2}{2m_{B_1}} h_i. \end{aligned} $$
Moreover, with the same computations as in Lemma 3, for all $nh_i \in [\bar{t} - \varepsilon, \bar{t} + \varepsilon] \cap [0, \tau]$ such that $n \ge 1$ we have

$$M(U^n)(V^{n-1} - V^n + h_i F^n) = \sum_{\beta \in J(U^{n+1})} \mu_{\beta}^{n} M^{1/2}(U^{n+1}) v_{\beta}(U^{n+1}) \quad (41)$$

with

$$-3 \frac{\lambda_{\max, B_1}}{\sqrt{\lambda_{\min, B_1}}} C_0 C_{*, \bar{u}} = -C'_2 \le \mu_{\beta}^n \le 0 \quad \forall \beta \in J(U^{n+1}). \quad (42)$$

Thus, for all $h_i \in (0, h_\varepsilon]$ we get

$$\begin{aligned} & \left(M^{1/2}(U^{p_i+1})V^{p_i}, v_{\alpha}(U^{p_i+1})\right) \\ &= \left(M^{1/2}(U^{N_i+1})V^{N_i}, v_{\alpha}(U^{N_i+1})\right) \\ &\quad + \left(M^{1/2}(U^{N_i+1})V^{N_i}, v_{\alpha}(U^{p_i+1}) - v_{\alpha}(U^{N_i+1})\right) \\ &\quad + \sum_{n=N_i+1}^{p_i} \left(M^{1/2}(U^{n+1})V^n - M^{1/2}(U^n)V^{n-1}, v_{\alpha}(U^{p_i+1})\right) \\ &\le \frac{L_f C_0^2}{2m_{B_1}} h_i + \left(M^{1/2}(U^{N_i+1})V^{N_i}, v_{\alpha}(U^{p_i+1}) - v_{\alpha}(U^{N_i+1})\right) \\ &\quad + \sum_{n=N_i+1}^{p_i} \left(\left(M^{1/2}(U^{n+1}) - M^{1/2}(U^n)\right)V^{n-1}, v_{\alpha}(U^{p_i+1})\right) \\ &\quad + \sum_{n=N_i+1}^{p_i} h_i \left(M^{1/2}(U^{n+1})F^n, v_{\alpha}(U^{p_i+1})\right) \\ &\quad + \sum_{n=N_i+1}^{p_i} \left(M^{1/2}(U^{n+1})(V^n - V^{n-1} - h_i F^n), v_{\alpha}(U^{p_i+1})\right). \end{aligned}$$

Using the Lipschitz property of $M^{1/2}$ on $B_1$ and recalling that the mappings $v_\alpha$ ($\alpha \in \{1, \dots, \nu\}$) are $L_{\bar{u}}$-Lipschitz continuous on $\bar{B}(\bar{u}, r_{\bar{u}})$, we get

$$\begin{aligned} & (M^{1/2}(U^{p_i+1})V^{p_i}, v_{\alpha}(U^{p_i+1})) \\ &\le \sum_{n=N_i+1}^{p_i} h_i \left(L_{M^{1/2}} C_0^2 + \sqrt{\lambda_{\max, B_1}} C_F\right) + \frac{L_f C_0^2}{2m_{B_1}} h_i \\ &\quad + 2\varepsilon \sqrt{\lambda_{\max, B_1}} C_0^2 L_{\bar{u}} + \sum_{n=N_i+1}^{p_i} \left(M^{1/2}(U^{n+1})(V^n - V^{n-1} - h_i F^n), v_{\alpha}(U^{p_i+1})\right). \end{aligned} \quad (43)$$
There remains the task of estimating the last term. Using (41) and (42) we rewrite it as follows

$$
\begin{align}
& \sum_{n=N_i+1}^{p_i} \sum_{\beta \in J(U^{n+1})} (-\mu_{\beta}^n) (M^{1/2}(U^{n+1})M^{-1}(U^n)M^{1/2}(U^{n+1})v_{\beta}(U^{n+1}), v_{\alpha}(U^{p_i+1})) \nonumber \\
& \le \sum_{n=N_i+1}^{p_i} \sum_{\beta \in J(U^{n+1})} C_2' \|M^{1/2}(U^{n+1})\|^2 \|M^{-1}(U^{n+1}) - M^{-1}(U^n)\| \nonumber \\
& \quad + \sum_{n=N_i+1}^{p_i} \sum_{\beta \in J(U^{n+1})} (-\mu_{\beta}^n) (v_{\beta}(U^{n+1}), v_{\alpha}(U^{p_i+1})). \tag{44}
\end{align}
$$

By definition of $N_i$ we have $\alpha \notin J(U^{n+1})$ for all $n \in \{N_i+1, \dots, p_i\}$. Moreover, from assumption (H6) we have

$$
(v_{\beta}(\bar{u}), v_{\alpha}(\bar{u})) = (e_{\beta}(\bar{u}), e_{\alpha}(\bar{u})) \le 0 \quad \forall \beta \in J(\bar{u}) \setminus \{\alpha\}
$$

and (35) and (36) imply that $J(U^{n+1}) \subset J(\bar{u})$ for all $nh_i \in [\bar{t} - \varepsilon, \bar{t} + \varepsilon] \cap [0, \tau]$.
It follows that

$$
\begin{align*}
& \sum_{n=N_i+1}^{p_i} \sum_{\beta \in J(U^{n+1})} (-\mu_{\beta}^n) (v_{\beta}(U^{n+1}), v_{\alpha}(U^{p_i+1})) \\
\le & \sum_{n=N_i+1}^{p_i} \sum_{\beta \in J(U^{n+1})} (-\mu_{\beta}^n) \left( (v_{\beta}(U^{n+1}), v_{\alpha}(U^{p_i+1})) - (v_{\beta}(\bar{u}), v_{\alpha}(\bar{u})) \right) \\
\le & \sum_{n=N_i+1}^{p_i} \sum_{\beta \in J(U^{n+1})} |\mu_{\beta}^n| L_{\bar{u}} \left( |U^{n+1} - \bar{u}| + |U^{p_i+1} - \bar{u}| \right).
\end{align*}
$$

Hence,

$$
\begin{align*}
& \sum_{n=N_i+1}^{p_i} \sum_{\beta \in J(U^{n+1})} (-\mu_{\beta}^n) (M^{1/2}(U^{n+1})M^{-1}(U^n)M^{1/2}(U^{n+1})v_{\beta}(U^{n+1}), v_{\alpha}(U^{p_i+1})) \\
& \le \sum_{n=N_i+1}^{p_i} \sum_{\beta \in J(U^{n+1})} C_2' \lambda_{max,B_1} L_{M^{-1}} C_0 h_i \\
& \quad + \sum_{n=N_i+1}^{p_i} \sum_{\beta \in J(U^{n+1})} 2 |\mu_{\beta}^n| L_{\bar{u}} (\|u - u_{h_i}\|_{C^0([0,\tau];\mathbb{R}^d)} + C_0(\varepsilon+h_i)).
\end{align*}
$$

But

$$
\begin{align*}
|\mu_{\beta}^{n}| &= \left| (M^{-1/2}(U^{n+1})M(U^n)(V^{n-1} - V^n + h_i F^n), w_{\beta}(U^{n+1})) \right| \\
&\le \frac{\lambda_{\max,B_1}}{\sqrt{\lambda_{\min,B_1}}} (|V^{n-1} - V^n| + h_i C_F) C_{*,\bar{u}}
\end{align*}
$$
for all $\beta \in J(U^{n+1})$, for all $n \in \{N_i + 1, \dots, p_i\}$. Hence, with the estimate of the discrete accelerations obtained at Proposition 3

$$ 
\begin{aligned}
& \sum_{n=N_i+1}^{p_i} \sum_{\beta \in J(U^{n+1})} (-\mu_{\beta}^n) (M^{1/2}(U^{n+1})M^{-1}(U^n)M^{1/2}(U^{n+1})v_{\beta}(U^{n+1}), v_{\alpha}(U^{p_i+1})) \\
& \le 2\nu \frac{\lambda_{\max,B_1}}{\sqrt{\lambda_{\min,B_1}}} C_{*,\bar{u}} (C_0' + 2\epsilon C_F) L_{\bar{u}} \left( \|u - u_{h_i}\|_{C^0([0,\tau];\mathbb{R}^d)} + C_0(\epsilon + h_i) \right) \\
& \quad + 2\epsilon v C_2' \lambda_{\max,B_1} L_{M^{-1}} C_0.
\end{aligned} 
\quad (45) $$

Finally, combining (43), (44) and (45), we obtain

$$ 
\begin{aligned}
& (M^{1/2}(U^{p_i+1})V^{p_i}, v_{\alpha}(U^{p_i+1})) \\
& \le \frac{L_f C_0^2}{2m_{B_1}} h_i + 2\epsilon C_0^2 \sqrt{\lambda_{\max,B_1}} L_{\bar{u}} \\
& \quad + 2\epsilon (L_{M^{1/2}} C_0^2 + \sqrt{\lambda_{\max,B_1}} C_F + \nu C_2' \lambda_{\max,B_1} L_{M^{-1}} C_0) \\
& \quad + 2\nu \frac{\lambda_{\max,B_1}}{\sqrt{\lambda_{\min,B_1}}} C_{*,\bar{u}} (C_0' + 2\epsilon C_F) L_{\bar{u}} \left( \|u - u_{h_i}\|_{C^0([0,\tau];\mathbb{R}^d)} + C_0(\epsilon + h_i) \right)
\end{aligned} 
$$

for all $h_i \in (0, h_\epsilon]$, for all $\epsilon \in (0, \bar{\epsilon}]$, which proves (40). Passing to the limit as $h_i$ tends to zero, then when $\epsilon$ tends to zero, we may conclude the proof. $\square$

### 3.3. Study of the initial conditions

We can now prove quite easily that property (P4) is satisfied.

**Lemma 5.** The initial conditions $(u_0, v_0)$ are satisfied in the following sense:

$$ u(0) = u_0, \quad \dot{u}(0+0) = v_0. $$

**Proof.** Since the sequence $(u_{h_i})_{i \in \mathbb{N}}$ converges uniformly to $u$ on $[0, \tau]$, we have

$$ u(0) = \lim_{h_i \to 0} u_{h_i}(0). $$

But $u_{h_i}(0) = U^0 = u_0$ for all $h_i \in (0, h^*]$, and thus $u(0) = u_0$. From the results of the previous subsection we already know that

$$ \dot{u}(0+0) = \mathrm{Proj}_{M(u(0))} (T_K(u(0)), \dot{u}(0-0)) $$

where $\dot{u}(0-0) = \dot{u}(0) = v(0)$ (see (24)). It follows that

$$ \dot{u}(0+0) = \mathrm{Proj}_{M(u_0)} (T_K(u_0), v(0)). \quad (46) $$

Since the sequence $(v_{h_i})_{i \in \mathbb{N}}$ converges pointwise to $v$ on $[0, \tau]$ we have

$$ v(0) = \lim_{h_i \to 0} v_{h_i}(0). $$
Let us prove now that $\lim_{h_i \to 0} v_{h_i}(0) = v_0$. For all $h_i \in (0, h^*]$ we have

$$v_{h_i}(0) = V^0 = \frac{U^1 - U^0}{h_i}$$

and the definition of $U^1$ implies that

$$h_i \|v_0 + z(h_i) - v_{h_i}(0)\|_{M(u_0)} \le \|u_0 + h_i v_0 + h_i z(h_i) - Z\|_{M(u_0)} \quad \forall Z \in K,$$

which yields

$$\|v_0 - v_{h_i}(0)\|_{M(u_0)} \le 2 \|z(h_i)\|_{M(u_0)} + \|v_0 - v\|_{M(u_0)}$$

for all $v \in \mathbb{R}^d$ such that $u_0 + h_i v \in K$.

If $u_0 \in \text{Int}(K)$, we infer that there exists $r_{u_0} > 0$ such that $u_0 + h_i v_0 \in K$ for all $h_i \in (0, \frac{r_{u_0}}{|v_0|+1}]$ and thus

$$\|v_0 - v_{h_i}(0)\|_{M(u_0)} \le 2 \|z(h_i)\|_{M(u_0)} \quad \forall h_i \in \left(0, \min\left(\frac{r_{u_0}}{|v_0|+1}, h^*\right)\right].$$

Since $\lim_{h_i \to 0} z(h) = 0$, we get $\lim_{h_i \to 0} v_{h_i}(0) = v(0) = v_0$.

Let us assume now that $u_0 \in \partial K$. Since $v_0 \in T_K(u_0)$ and $\tilde{T}_K(u_0)$ is dense in $T_K(u_0)$, we may consider a sequence $(v_p)_{p \in \mathbb{N}^*}$ converging to $v_0$ and such that

$$v_p \in \tilde{T}_K(u_0) = \{w \in \mathbb{R}^d; (\nabla f_\alpha(u_0), w) > 0 \quad \forall \alpha \in J(u_0)\} \quad \forall p \in \mathbb{N}^*.$$

It follows that $(|v_p|)_{p \in \mathbb{N}^*}$ remains bounded, and we consider $M \in \mathbb{R}_+^*$ such that $M \ge |v_p|$ for all $p \in \mathbb{N}^*$. Using the continuity of the mappings $f_\alpha, \alpha \in \{1, \dots, v\}$, there exists $r_{u_0} > 0$ such that

$$f_{\alpha}(q) \ge \frac{f_{\alpha}(u_0)}{2} \quad \forall \alpha \notin J(u_0), \quad \forall q \in \bar{B}(u_0, r_{u_0}).$$

Let $p \in \mathbb{N}^*$. From the definition of $r_{u_0}$, we infer that

$$f_{\alpha}(u_0 + tv_p) \ge 0 \quad \forall \alpha \notin J(u_0), \quad \forall t \in (0, \frac{r_{u_0}}{M}).$$

Moreover, if $\alpha \in J(u_0)$

$$\begin{aligned} f_{\alpha}(u_0 + tv_p) &= f_{\alpha}(u_0) + t(\nabla f_{\alpha}(u_0), v_p) + t \int_0^1 (\nabla f_{\alpha}(u_0 + stv_p) - \nabla f_{\alpha}(u_0), v_p) \, ds. \end{aligned}$$

If $t \in (0, 1/M]$, we get $|tv_p| \le 1$ and thus $u_0 + stv_p \in B_1$ for all $s \in [0, 1]$, which yields

$$f_{\alpha}(u_0 + tv_p) \ge t (\nabla f_{\alpha}(u_0), v_p) - t^2 \frac{L_f |v_p|^2}{2} \quad \forall \alpha \in J(u_0).$$
It follows that there exists $t_p \in (0, \min(1, r_{u_0})/M)$ such that $f_\alpha(u_0 + tv_p) \ge 0$ for all $\alpha \in \{1, \dots, v\}$ and for all $t \in (0, t_p]$. Thus

$$ \begin{gathered} \|v_0 - v_{h_i}(0)\|_{M(u_0)} \le 2 \|z(h_i)\|_{M(u_0)} + \|v_0 - v_p\|_{M(u_0)} \\ \forall h_i \in (0, \min(h^*, t_p)], \quad \forall p \in \mathbb{N}^*. \end{gathered} $$

Then, passing to the limit when $h_i$ tends to zero, we get

$$ \|v_0 - v(0)\|_{M(u_0)} \le \|v_0 - v_p\|_{M(u_0)} \quad \forall p \in \mathbb{N}^* $$

and recalling that the sequence $(v_p)_{p \in \mathbb{N}^*}$ converges to $v_0$, we obtain $v_0 = v(0)$. Finally, using (46) and recalling that $v_0 \in T_K(u_0)$, we get

$$ \dot{u}(0+0) = \mathrm{Proj}_{M(u_0)}(T_K(u_0), v_0) = v_0. $$

□

With the previous results, we can state the following theorem:

**Theorem 2.** Let us assume that there exist $C_0 > 0$, $\tau_0 > 0$, $h_0^* \in (0, h^*]$ and a subsequence of the approximate positions defined by (5)–(7) such that

$$ |V^n| = \left| \frac{U^{n+1} - U^n}{h_i} \right| \le C_0 \quad \forall nh_i \in [0, \min(\tau_0, T)], \quad \forall h_i \in (0, h_0^*] $$

with $(h_i)_{i \in \mathbb{N}}$ decreasing to zero. Let $u_h$ and $v_h$ be defined by (19) and (20). Then, there exist a subsequence still denoted $(h_i)_{i \in \mathbb{N}}$ and $(u, v) \in C^0([0, \min(\tau_0, T)]; \mathbb{R}^d) \times BV(0, \min(\tau_0, T); \mathbb{R}^d)$ such that

$$ \begin{gathered} u_{h_i} \to u \quad \text{strongly in } C^0([0, \min(\tau_0, T)]; \mathbb{R}^d), \\ v_{h_i} \to v \quad \text{pointwise in } [0, \min(\tau_0, T)], \end{gathered} $$

with

$$ u(t) = u_0 + \int_0^t v(s) \, ds \quad \text{for all } t \in [0, \min(\tau_0, T)], $$

and $u$ is a solution of problem (P) on $[0, \min(\tau_0, T)]$.

By combining Theorem 2 with the a priori estimate of the discrete velocities obtained in Theorem 1, we immediately obtain a local convergence result for the numerical scheme, and thus a local existence result for problem (P).
### 4. Energy estimates and global results

In order to establish global convergence results, we now state an energy estimate for the solutions of problem (P).

**Proposition 6.** Let $C > \|v_0\|_{M(u_0)}$. Then there exists $\tau(C) > 0$ such that, for any solution $u$ of problem (P) defined on $[0, \tau]$ (with $\tau \in (0, T]$), we have

$$ 
\begin{gathered}
|u(t) - u_0| \le C \quad \forall t \in [0, \min(\tau(C), \tau)], \\
\|\dot{u}(t)\|_{M(u(t))} \le C \quad \text{dt almost everywhere on } [0, \min(\tau(C), \tau)].
\end{gathered}
$$

**Proof.** Let us define the kinetic energy $E$ by

$$ E = \frac{1}{2} (\dot{u}, M(u)\dot{u}). $$

Since $\dot{u} \in BV(0, \tau; \mathbb{R}^d)$ and $u$ is absolutely continuous from $[0, \tau]$ to $\mathbb{R}^d$, $E$ belongs to $BV(0, \tau; \mathbb{R}^d)$. Moreover (see [8])

$$ dE = \left( d\dot{u}, M(u) \left( \frac{\dot{u}^+ + \dot{u}^-}{2} \right) \right) + \frac{1}{2} (\dot{u}, (dM(u)\dot{u})\dot{u}) dt. $$

Let $[t_1, t_2] \subset [0, \tau)$. Then

$$ 
\begin{aligned}
E(t_2 + 0) - E(t_1 + 0) = & \int_{(t_1, t_2]} \left( d\dot{u}, M(u) \left( \frac{\dot{u}^+ + \dot{u}^-}{2} \right) \right) \\
& + \frac{1}{2} \int_{t_1}^{t_2} (\dot{u}(t), (dM(u(t))\dot{u}(t))\dot{u}(t)) dt.
\end{aligned}
$$

Let us define $D = \{t \in (t_1, t_2]; \dot{u}(t+0) \neq \dot{u}(t-0)\}$. The set $D$ is at most denumerable and

$$ 
\int_{(t_1, t_2]} \left( d\dot{u}, M(u) \left( \frac{\dot{u}^+ + \dot{u}^-}{2} \right) \right) = \int_{(t_1, t_2] \setminus D} \left( d\dot{u}, M(u)\dot{u}^+ \right) \\
+ \sum_{t \in D} \frac{1}{2} \left( \|\dot{u}(t+0)\|_{M(u(t))}^2 - \|\dot{u}(t-0)\|_{M(u(t))}^2 \right).
$$

But, with property (P2), we have also

$$ 
\begin{gathered}
M(u(t)) v'_{\mu}(t) - g(t, u(t), \dot{u}(t)) t'_{\mu}(t) \in -N_K(u(t)) \\
d\mu \text{ almost everywhere on } [0, \tau]
\end{gathered}
$$

where $d\mu = |d\dot{u}| + dt$ and $v'_{\mu}$ and $t'_{\mu}$ are, respectively, the densities of $d\dot{u}$ and $dt$ with respect to $\mu$. Thus

$$ 
\int_{(t_1, t_2] \setminus D} (d\dot{u}, M(u)\dot{u}^+) = \int_{(t_1, t_2] \setminus D} (g(t, u, \dot{u}), \dot{u}^+) t'_{\mu} d\mu \\
+ \int_{(t_1, t_2] \setminus D} (M(u)v'_{\mu} - g(t, u, \dot{u})t'_{\mu}, \dot{u}^+) d\mu.
$$
Since $\dot{u}(t+0) = \dot{u}(t-0) \in (T_K(u(t))) \cap (-T_K(u(t)))$ for all $t \in (t_1, t_2] \setminus D$, the last term vanishes and we get

$$ 
\begin{aligned}
E(t_2 + 0) - E(t_1 + 0) ={}& \frac{1}{2} \int_{t_1}^{t_2} (\dot{u}(t), (dM(u(t))\dot{u}(t))\dot{u}(t)) \, dt \\
& + \int_{(t_1, t_2] \setminus D} (g(t, u, \dot{u}), \dot{u}^+) t'_{\mu} d\mu \\
& + \sum_{t \in D} \frac{1}{2} \left( \|\dot{u}(t+0)\|_{M(u(t))}^2 - \|\dot{u}(t-0)\|_{M(u(t))}^2 \right).
\end{aligned}
$$

But, with property (P3), we know that

$$ 2E(t+0) = \|\dot{u}(t+0)\|_{M(u(t))}^2 \le \|\dot{u}(t-0)\|_{M(u(t))}^2 = 2E(t-0) \quad \forall t \in (0, \tau) $$

and finally

$$ E(t_2 + 0) - E(t_1 + 0) \le \int_{t_1}^{t_2} (g(t, u(t), \dot{u}(t)), \dot{u}(t)) \, dt \\ + \frac{1}{2} \int_{t_1}^{t_2} (\dot{u}(t), (dM(u(t))\dot{u}(t))\dot{u}(t)) \, dt. $$

In particular, for all $t \in [0, \tau)$

$$ 
\begin{aligned}
E(t+0) \le E(0+0) &+ \int_0^t (g(s, u(s), \dot{u}(s)), \dot{u}(s)) \, ds \\
&+ \frac{1}{2} \int_0^t (\dot{u}(s), (dM(u(s))\dot{u}(s))\dot{u}(s)) \, ds.
\end{aligned}
$$

Observing that $C^2 > 2E(0+0)$, the continuity of $u$ on $[0, \tau]$ and the right continuity of $E(\cdot + 0)$ on $[0, \tau)$ imply that there exists $\bar{\tau} \in (0, \tau)$ such that

$$ |u(t) - u(0)| \le C, \quad E(t+0) \le \frac{C^2}{2} \quad \forall t \in [0, \bar{\tau}]. \quad (47) $$

We define

$$ \tau_{max} = \sup \{ \bar{\tau} \in (0, \tau) \text{ such that (47) holds} \}. $$

Since $u$ is continuous on $[0, \tau]$, we have

$$ |u(t) - u(0)| \le C \quad \forall t \in [0, \tau_{max}], \quad (48) $$

and

$$ E(t+0) = \frac{1}{2} \left| M^{1/2}(u(t)) \dot{u}(t+0) \right|^2 \le \frac{C^2}{2} \quad \forall t \in [0, \tau_{max}). \quad (49) $$

It follows that

$$ \|\dot{u}(t)\|_{M(u(t))}^2 = \left| M^{1/2}(u(t)) \dot{u}(t) \right|^2 \le C^2 dt \quad \text{almost everywhere on } [0, \tau_{max}]. $$
If $\tau_{\max} = \tau$ there is nothing to prove. Otherwise, we define

$$ \alpha = \sup \left\{ \left| (g(t, q, M^{-1/2}(q)w), M^{-1/2}(q)w) \right| ; t \in [0, T], q \in \bar{B}(u_0, C), w \in \bar{B}(0, C) \right\}, $$

$$ \beta = \sup \left\{ \frac{1}{2} \left\| M^{-1/2}(q) (dM(q)M^{-1/2}(q)w) M^{-1/2}(q) \right\|, q \in \bar{B}(u_0, C), w \in \bar{B}(0, C) \right\} $$

$$ \gamma = \sup \{ \| M^{-1/2}(q) \|, q \in \bar{B}(u_0, C) \} $$

With (48) and (49) we obtain that for all $t \in [0, \tau_{\max}]$

$$ E(t+0) - E(0+0) \le \int_0^t \left( \alpha + \beta \left| M^{1/2}(u(s)) \dot{u}(s) \right|^2 \right) ds \le (\alpha + \beta C^2)t $$

and

$$ |u(t) - u_0| \le \int_0^t |\dot{u}(s)| ds \le \int_0^t \gamma |M^{1/2}(u(s)) \dot{u}(s)| ds \le \gamma Ct. $$

Then, the continuity of $u$ on $[0, \tau]$ and the right continuity of $E(\cdot + 0)$ on $[0, \tau)$ imply that $\tau_{\max} \ge \min(\tau, \tau(C))$ where $\tau(C)$ is defined by

$$ \tau(C) = \begin{cases} \min\left(\frac{1}{\gamma}, \frac{C^2 - 2E(0+0)}{2(\alpha + \beta C^2)}\right) & \text{if } \alpha \ne 0 \text{ or } \beta \ne 0, \\ \frac{1}{\gamma} & \text{if } \alpha = 0 \text{ and } \beta = 0. \end{cases} $$

□

Now we can prove that

**Theorem 3.** Let $C > \|v_0\|_{M(u_0)}$ and $\tau(C) > 0$ such that, for any solution $u$ of problem (P) defined on $[0, \tau]$ (with $\tau \in (0, T]$), we have

$$ \begin{gathered} |u(t) - u_0| \le C \quad \forall t \in [0, \min(\tau(C), \tau)], \\ \|\dot{u}(t)\|_{M(u(t))} \le C \quad \text{dt almost everywhere on } [0, \min(\tau(C), \tau)]. \end{gathered} $$

Let $u_h$ and $v_h$ be the approximate positions and velocities defined by (19) and (20). Then, there exists a subsequence $(u_{h_i}, v_{h_i})_{i \in \mathbb{N}}$, $\tau \in [\min(\tau(C), T), T]$ and $(u, v) \in C^0([0, \tau]; \mathbb{R}^d) \times BV(0, \tau; \mathbb{R}^d)$ such that

$$ u_{h_i} \to u \quad \text{strongly in } C^0([0, \tau]; \mathbb{R}^d), $$

$$ v_{h_i} \to v \quad \text{pointwise in } [0, \tau], $$

with

$$ u(t) = u_0 + \int_0^t v(s) ds \quad \forall t \in [0, \tau] $$

and $u$ is a solution of problem (P) on $[0, \tau]$.
**Proof.** Let $C > \|v_0\|_{M(u_0)}$. We define $\mathcal{B} = \bar{B}(u_0, C+1)$ and $B_0, \lambda_{\min}, \lambda_{\max}$ by (9) and (10), respectively. Let us choose $C_0$ such that

$$C_0 > \sqrt{\frac{\lambda_{\max}}{\lambda_{\min}}} C_0^*$$

with

$$C_0^* = \max \left( 2\sqrt{\frac{\lambda_{\max}}{\lambda_{\min}}} (|v_0| + 1), C' \right) \\ C' = (C + 1) \sup \left\{ \|M^{-1/2}(q)\| ; q \in \bar{B}(u_0, C+1) \right\}$$

and let $C_F$ be defined by (11).

Then, from Theorem 1, we know that there exists $h_0^* \in (0, h^*]$ and $\tau_0 > 0$, depending only on $\mathcal{B}, C_0, C_0^*$ and the data, such that

$$|V^n| = \left| \frac{U^{n+1} - U^n}{h} \right| \le C_0 \quad \forall nh \in [0, \min(\tau_0, T)], \quad \forall h \in (0, h_0^*]. \quad (50)$$

Moreover, from Proposition 2, we know also that, for all $t_{0h} \in [0, T)$ and for all $(\hat{U}^0, \hat{U}^1) \in (\mathcal{B} \cap K) \times K$ such that

$$|\hat{U}^1 - \hat{U}^0| \le h C_0^* \quad \forall h \in (0, h_0^*]$$

the approximate positions defined by

$$\hat{U}^{n+1} \in \operatorname{Argmin}_{Z \in K} \|\hat{W}^n - Z\|_{M(\hat{U}^n)}$$

with

$$\hat{W}^n = 2\hat{U}^n - \hat{U}^{n-1} + h^2 \hat{F}^n, \quad \hat{F}^n = F \left( t_{0h} + nh, \hat{U}^n, \frac{\hat{U}^n - \hat{U}^{n-1}}{h}, h \right)$$

for all $n \in \{1, \dots, \lfloor \frac{T-t_{0h}}{h} \rfloor\}$ and for all $h \in (0, h_0^*]$, satisfy

$$\left| \frac{\hat{U}^{n+1} - \hat{U}^n}{h} \right| \le C_0 \quad \forall nh \in [0, \min(\tau_0, T - t_{0h})], \quad \forall h \in (0, h_0^*].$$

Let $\tau(h) = m(h)h$ be the maximal discrete time step such that estimate (50) holds, that is, for all $h \in (0, h_0^*]$

$$m(h) = \max \left\{ n \in \{0, \dots, \lfloor T/h \rfloor\} ; |V^k| \le C_0 \quad \forall k \in \{0, \dots, n\} \right\}.$$

We define $\tau_1 = \liminf_{h \to 0} \tau(h) = \liminf_{h \to 0} m(h)h$. Theorem 1 implies that $\tau_1 \ge \tau' = \min(\tau_0, T)$. Let us now distinguish two subcases.
Case 1: $\tau(C) < T$.

Let us prove that $\tau_1 > \tau(C) = \min(\tau(C), T)$. Indeed, assume that $\tau_1 \le \tau(C)$ and let $\varepsilon \in (0, \tau'/8)$. Then, there exists a subsequence $(h_i)_{i \in \mathbb{N}}$, decreasing to zero, such that $(\tau(h_i))_{i \in \mathbb{N}}$ converges to $\tau_1$ and there exists $h_\varepsilon^* \in (0, \min(h_0^*, \tau'/8)]$ such that $m(h_i)h_i \ge \tau_1 - \varepsilon$ for all $h_i \in (0, h_\varepsilon^*]$. We may apply theorem 2 with $h_0^* := h_\varepsilon^*$ and $\tau_0 := \tau_1 - \varepsilon$; we infer that there exists a subsequence, still denoted $(h_i)_{i \in \mathbb{N}}$, such that $(u_{h_i}, v_{h_i})_{i \in \mathbb{N}}$ converges to a solution of problem (P) on $[0, \tau_1 - \varepsilon]$. Thus, with Proposition 6 we get

$$|u(t) - u_0| \le C \quad \forall t \in [0, \tau_1 - \varepsilon], \\ \|\dot{u}(t)\|_{M(u(t))} \le C \quad \text{dt almost everywhere on } [0, \tau_1 - \varepsilon].$$

Now we prove that:

**Lemma 6.** We have

$$\limsup_{h_i \to 0^+} \sup_{t \to 0^+} \left\{ \|V^n\|_{M(U^n)} , 0 \le nh_i \le \tau_1 - \varepsilon \right\} \\ \le \ess \sup \left\{ \|\dot{u}(t)\|_{M(u(t))} , 0 \le t \le \tau_1 - \varepsilon \right\}.$$

**Proof.** Let us prove this result by contradiction. Assume that

$$\limsup_{h_i \to 0^+} \sup_{t \to 0^+} \left\{ \|V^n\|_{M(U^n)} , 0 \le nh_i \le \tau_1 - \varepsilon \right\} > S$$

with $S = \ess \sup \{\|\dot{u}(t)\|_{M(u(t))} , 0 \le t \le \tau_1 - \varepsilon\}$. Then, there exist $\gamma > 0$, $\tilde{h}_\varepsilon^* \in (0, h_\varepsilon^*]$ and a subsequence $(h_{\varphi(i)})_{i \in \mathbb{N}}$ decreasing to zero such that

$$\sup \left\{ \|V^n\|_{M(U^n)} , 0 \le nh_{\varphi(i)} \le \tau_1 - \varepsilon \right\} \ge S + \gamma \quad \forall h_{\varphi(i)} \in (0, \tilde{h}_\varepsilon^*].$$

It follows that there exists $n_{\varphi(i)} \in \{0, \dots, \lfloor(\tau_1 - \varepsilon)/h_{\varphi(i)}\rfloor\}$ such that

$$\|V^{n_{\varphi(i)}}\|_{M(U^{n_{\varphi(i)}})} \ge S + \gamma \quad \forall h_{\varphi(i)} \in (0, \tilde{h}_\varepsilon^*].$$

Possibly extracting another subsequence, still denoted $(h_{\varphi(i)})_{i \in \mathbb{N}}$, we may assume without loss of generality that the sequence $(n_{\varphi(i)}h_{\varphi(i)})_{i \in \mathbb{N}}$ converges to a limit $\tilde{\tau} \in [0, \tau_1 - \varepsilon]$.

First, we observe that $\tilde{\tau} > 0$. Indeed, with the same computations as in Proposition 2 (see (13)), we obtain that, for all $h_i \in (0, h_\varepsilon^*]$ and for all $nh_i \in [0, \tau_1 - \varepsilon]$

$$\|V^n\|_{M(U^n)} \le \|V^{n-1}\|_{M(U^{n-1})} + C_2h_i \le \|V^0\|_{M(U^0)} + C_2nh_i \quad (51)$$

with

$$C_2 = \sqrt{\lambda_{\max, B_1}} C_F + L_{M^{1/2}} C_0^2 + \frac{3L_f}{2m_{B_1}} \frac{\lambda_{\max, B_1}}{\lambda_{\min, B_1}} \nu C_{*, B_1} C_0^2,$$

$B_1 = \bar{B}(u_0, C_0T + 1)$ and $\lambda_{\max, B_1}, \lambda_{\min, B_1}$ given by (15).
Thus, for all $h_{\varphi(i)} \in (0, \tilde{h}_{\varepsilon}^*]$

$$S + \gamma \leq \|V^{n_{\varphi(i)}}\|_{M(U^{n_{\varphi(i)}})} \leq \|v_{h_{\varphi(i)}}(0)\|_{M(u_0)} + C_2 n_{\varphi(i)} h_{\varphi(i)}$$

and at the limit when $i$ tends to $+\infty$, we get

$$S + \gamma \leq \|v(0)\|_{M(u_0)} + C_2 \tilde{\tau} = \|v_0\|_{M(u_0)} + C_2 \tilde{\tau}.$$

On the other hand, the right continuity of $\|\dot{u}(\cdot + 0)\|_{M(u)}$ implies that, for all $\rho > 0$, there exists $\tau_{\rho} \in (0, \tau_1 - \varepsilon]$ such that, for all $t \in [0, \tau_{\rho}]$

$$\left\| \dot{u}(t+0) \right\|_{M(u(t))} - \left\| \dot{u}(0+0) \right\|_{M(u_0)} \leq \rho.$$

It follows that

$$\|\dot{u}(0+0)\|_{M(u_0)} - \rho = \|v_0\|_{M(u_0)} - \rho \leq \|\dot{u}(t)\|_{M(u(t))} \quad \text{almost everywhere on } [0, \tau_{\rho}],$$

and thus

$$\|v_0\|_{M(u_0)} - \rho \leq S \quad \forall \rho > 0.$$

Hence $\|v_0\|_{M(u_0)} \leq S$ and $\tilde{\tau} > 0$.

Then, once again using the estimate (51), we obtain

$$S + \gamma \leq \|V^{n_{\varphi(i)}}\|_{M(U^{n_{\varphi(i)}})} \leq \|V^{n_{\varphi(i)}-p}\|_{M(U^{n_{\varphi(i)}-p})} + C_2 p h_{\varphi(i)}$$

for all $p \in \{0, \dots, n_{\varphi(i)}\}$, for all $h_{\varphi(i)} \in (0, \tilde{h}_{\varepsilon}^*]$. It follows that

$$S + \frac{\gamma}{2} \leq \|V^k\|_{M(U^k)}$$

for all $kh_{\varphi(i)} \in [\max(0, n_{\varphi(i)}h_{\varphi(i)} - \frac{\gamma}{2C_2}), n_{\varphi(i)}h_{\varphi(i)}]$. Moreover, for all $t \in [kh_{\varphi(i)}, (k+1)h_{\varphi(i)}),$

$$\begin{aligned} \left\| v_{h_{\varphi(i)}}(t) \right\|_{M(U^k)} - \left\| v_{h_{\varphi(i)}}(t) \right\|_{M(u_{h_{\varphi(i)}}(t))} &\leq L_{M^{1/2}} \left| U^k - u_{h_{\varphi(i)}}(t) \right| C_0 \\ &\leq L_{M^{1/2}} C_0^2 h_{\varphi(i)}. \end{aligned}$$

Since $(n_{\varphi(i)}h_{\varphi(i)})_{i \in \mathbb{N}}$ converges to $\tilde{\tau} > 0$, we infer that there exists an interval $I \subset [0, \tau_1 - \varepsilon]$ with a non-empty interior, such that

$$I \subset \left[ \max \left( 0, (n_{\varphi(i)} + 1)h_{\varphi(i)} - \frac{\gamma}{2C_2} \right), n_{\varphi(i)}h_{\varphi(i)} \right]$$

and

$$S + \frac{\gamma}{4} \leq \|v_{h_{\varphi(i)}}(t)\|_{M(u_{h_{\varphi(i)}}(t))} \quad \forall t \in I$$

for all $h_{\varphi(i)}$ small enough.

Then, passing to the limit as $i$ tends to $+\infty$, we obtain

$$S + \frac{\gamma}{4} \leq \|v(t)\|_{M(u(t))} \quad \forall t \in I$$

But $v(t) = \dot{u}(t)$ almost everywhere on $I$, and

$$S \geq \operatorname*{ess\,sup}\{\|\dot{u}(t)\|_{M(u(t))}; t \in I\}$$

which yields a contradiction. □
With the previous lemma, possibly decreasing $h_{\epsilon}^*$ we get

$$U^n \in \bar{B}(u_0, C + 1), \quad \|V^n\|_{M(U^n)} \le C + 1 \quad \forall nh_i \in [0, \tau_1 - \varepsilon], \forall h_i \in (0, h_{\epsilon}^*].$$

It follows that

$$|V^n| \le (C + 1) \sup \left\{ \|M^{-1/2}(q)\| ; q \in \bar{B}(u_0, C + 1) \right\} = C' \le C_0^* < C_0.$$

We choose now $l(h_i) \in \{0, \dots, \lfloor T/h \rfloor\}$ such that

$$l(h_i)h_i \in \left[ \tau_1 - \frac{\tau'}{2}, \tau_1 - \frac{\tau'}{4} \right] \quad \forall h_i \in (0, h_{\epsilon}^*]$$

and let

$$\hat{U}^0 = U^{l(h_i)}, \quad \hat{U}^1 = U^{l(h_i)+1}, \quad t_{0h_i} = l(h_i)h_i.$$

We have

$$|\hat{U}^1 - \hat{U}^0| = h_i \left|V^{l(h_i)}\right| \le C_0^* h_i$$

and $\hat{U}^0, \hat{U}^1$ belong to $\mathcal{B} \cap K$ for all $h_i \in (0, h_{\epsilon}^*]$.

Then, for all $n \in \{l(h_i), \dots, \lfloor \frac{T}{h_i} \rfloor\}$, $U^n = \hat{U}^{n-l(h_i)}$ and with Proposition 2, we obtain

$$|V^n| = \left| \frac{\hat{U}^{n-l(h_i)+1} - \hat{U}^{n-l(h_i)}}{h_i} \right| \le C_0$$

for all $(n-l(h_i))h_i \in [0, \min(\tau_0, T-l(h_i)h_i)]$, for all $h_i \in (0, h_{\epsilon}^*]$.

Hence

$$m(h_i)h_i > l(h_i)h_i + \min(\tau_0, T-l(h_i)h_i) - h_i = \min(l(h_i)h_i + \tau_0, T) - h_i$$

for all $h_i \in (0, h_{\epsilon}^*]$. But $l(h_i)h_i \ge \tau_1 - \frac{\tau'}{2}$ and $\tau' = \min(\tau_0, T) = \tau_0$, so

$$m(h_i)h_i \ge \min \left( \tau_1 + \frac{\tau'}{2}, T \right) - h_i \quad \forall h_i \in (0, h_{\epsilon}^*]$$

and, at the limit, we get

$$\tau_1 = \lim_{h_i \to 0} m(h_i)h_i \ge \min \left( \tau_1 + \frac{\tau'}{2}, T \right)$$

which is absurd.

Thus $\tau_1 = \lim_{h_i \to 0} m(h_i)h_i > \tau(C)$ and there exists $\tilde{h}_0^* \in (0, h_0^*]$ such that

$$m(h_i)h_i = \tau(h_i) \ge \tau_1 - \frac{\tau_1 - \tau(C)}{2} = \frac{\tau_1 + \tau(C)}{2} > \tau(C) \quad \forall h_i \in (0, \tilde{h}_0^*].$$

Then we apply Theorem 2 with $\tau_0$ replaced by $\tau = \frac{\tau_1 + \tau(C)}{2}$ and $h_0^*$ by $\tilde{h}_0^*$, which yields the announced result.
Case 2: $\tau(C) \geq T$.

Since $m(h)h = \tau(h) \leq T$, we have $\tau_1 \leq T$. We consider once again $\varepsilon \in (0, \tau'/8)$, and we define as previously $h_{\varepsilon}^*$ and $l(h_i)$ for all $h_i \in (0, h_{\varepsilon}^*]$. Then, we have again

$$m(h_i)h_i > l(h_i)h_i + \min(\tau_0, T - l(h_i)h_i) - h_i \geq \min\left(\tau_1 + \frac{\tau'}{2}, T\right) - h_i \quad (52)$$

for all $h_i \in (0, h_{\varepsilon}^*]$. Thus, if $\tau_1 < T$, $\min(\tau_1 + \tau'/2, T) \in (\tau_1, T]$ and (52) yields a contradiction with the definition of $\tau_1 = \lim_{h_i \to 0} m(h_i)h_i$. We infer that $\tau_1 = T$ and (52) implies that $m(h_i)h_i > T - h_i$, that is, $m(h_i) \geq \lfloor T/h_i \rfloor$ for all $h_i \in (0, h_{\varepsilon}^*]$. Hence, we may apply Theorem 2 to obtain the convergence of a subsequence of $(u_{h_i}, v_{h_i})_{i \in \mathbb{N}}$, still denoted $(u_{h_i}, v_{h_i})_{i \in \mathbb{N}}$, to a solution of problem (P) on $[0, T]$. $\square$

# Appendix

**Lemma 7.** For all compact subset $\mathcal{B}$ of $\mathbb{R}^d$, there exist $m_{\mathcal{B}} > 0$ and $r_{\mathcal{B}} > 0$ such that for all $q \in K \cap \mathcal{B}$ and for all $\alpha \in J(q)$ we have

$$|M^{-1/2}(q')\nabla f_{\alpha}(q')| \geq m_{\mathcal{B}} \quad \forall q' \in \bar{B}(q, r_{\mathcal{B}}).$$

Furthermore, for all $q \in K \cap \mathcal{B}$, the family $(e_{\alpha}(q))_{\alpha \in J(q)}$ is linearly independent and can be completed as a basis $(v_j(q))_{1 \leq j \leq d}$. Let us denote by $(w_j(q))_{1 \leq j \leq d}$ the dual basis. Then there exists $C_{*,\mathcal{B}} > 0$ such that

$$|v_j(q)| = 1, \quad |w_j(q)| \leq C_{*,\mathcal{B}} \quad \forall j \in \{1, \dots, d\}, \quad \forall q \in K \cap \mathcal{B}.$$

**Proof.** Let $\mathcal{B}$ be a given compact subset of $\mathbb{R}^d$. For all $\alpha \in \{1, \dots, \nu\}$ we define

$$\mathcal{B}_{\alpha} = \mathcal{B} \cap \{q \in \mathbb{R}^d; f_{\alpha}(q) \leq 0\} \cap K.$$

Then $\mathcal{B}_{\alpha}$ is also a compact subset of $\mathbb{R}^d$ and, for all $q \in \mathcal{B}_{\alpha}$, we have $f_{\alpha}(q) = 0$. Hence, with (H2)

$$\nabla f_{\alpha}(q) \neq 0 \quad \forall q \in \mathcal{B}_{\alpha}, \quad \forall \alpha \in \{1, \dots, \nu\}.$$

It follows that there exists $m_{\alpha} > 0$ such that

$$m_{\alpha} = \inf_{q \in \mathcal{B}_{\alpha}} |M^{-1/2}(q) \nabla f_{\alpha}(q)|.$$

By continuity of the mappings $M^{-1/2}$ and $\nabla f_{\alpha}$, we infer that

$$\forall q \in \mathcal{B}_{\alpha}, \quad \exists \rho_q > 0 / \quad \left| M^{-1/2}(q') \nabla f_{\alpha}(q') \right| \geq \frac{m_{\alpha}}{2} \quad \forall q' \in \bar{B}(q, \rho_q).$$

Since $\mathcal{B}_{\alpha}$ is compact and $\mathcal{B}_{\alpha} \subset \bigcup_{q \in \mathcal{B}_{\alpha}} B(q, \frac{\rho_q}{2})$, there exists a finite set of points $\{q_1, \dots, q_p\} \in \mathcal{B}_{\alpha}^p$ such that $\mathcal{B}_{\alpha} = \bigcup_{i=1}^p B(q_i, \frac{\rho_{q_i}}{2})$.
By defining $\rho_{\alpha} = \min_{1 \le i \le p} \frac{\rho_{q_i}}{2}$ we obtain that

$$ \forall q \in \mathcal{B}_{\alpha}, \quad \left| M^{-1/2}(q') \nabla f_{\alpha}(q') \right| \ge \frac{m_{\alpha}}{2} \quad \forall q' \in \bar{B}(q, \rho_{\alpha}). $$

Finally, with

$$ m_{\mathcal{B}} = \min_{1 \le \alpha \le \nu} \frac{m_{\alpha}}{2}, \quad r_{\mathcal{B}} = \min_{1 \le \alpha \le \nu} \rho_{\alpha}, $$

we get the first part of the announced result.

As a consequence, for all $q \in K \cap \mathcal{B}$, we can define

$$ v_{\alpha}(q') = \frac{M^{-1/2}(q') \nabla f_{\alpha}(q')}{|M^{-1/2}(q') \nabla f_{\alpha}(q')|} \quad \forall q' \in \bar{B}(q, r_{\mathcal{B}}), \quad \forall \alpha \in J(q). $$

Let $q \in K \cap \mathcal{B}$ be given. From assumption (H3) we infer that $(e_{\alpha}(q))_{\alpha \in J(q)}$ is linearly independent, and there exists a family of vectors $(e_{\beta})_{\beta \in \{1, \dots, d\} \setminus J(q)}$ such that $|e_{\beta}| = 1$ for all $\beta \in \{1, \dots, d\} \setminus J(q)$ and $\{e_{\alpha}(q); \alpha \in J(q)\} \cup \{e_{\beta}; \beta \in \{1, \dots, d\} \setminus J(q)\}$ is a basis of $\mathbb{R}^d$.

Let us now define the mappings $v_{\beta}, \beta \in \{1, \dots, d\} \setminus J(q)$, by

$$ v_{\beta}(q') = e_{\beta} \quad \forall q' \in \bar{B}(q, r_{\mathcal{B}}). $$

The mappings $v_j, j \in \{1, \dots, d\}$, are continuous on $\bar{B}(q, r_{\mathcal{B}})$ and there exists $r_q \in (0, r_{\mathcal{B}}]$ such that $(v_j(q'))_{1 \le j \le d}$ is a basis of $\mathbb{R}^d$ for all $q' \in \bar{B}(q, r_q)$. Moreover, using the continuity of the mappings $f_{\alpha}, \alpha \in \{1, \dots, \nu\}$, and possibly decreasing $r_q$, we also have

$$ J(q') \subset J(q) \quad \forall q' \in \bar{B}(q, r_q). $$

It follows that

$$ v_{\alpha}(q') = e_{\alpha}(q') \quad \forall \alpha \in J(q'), \quad \forall q' \in \bar{B}(q, r_q) \cap K. $$

Let us denote by $(w_j(q'))_{1 \le j \le d}$ the dual basis of $(v_j(q'))_{1 \le j \le d}$ for all $q' \in \bar{B}(q, r_q)$. Then, the mappings $w_j, j \in \{1, \dots, d\}$, are continuous on $\bar{B}(q, r_q)$. Indeed, let $(\delta_j)_{1 \le j \le d}$ be the canonical basis of $\mathbb{R}^d$ and define $(a_{ij}(q'))_{1 \le i, j \le d}$ and $(b_{ij}(q'))_{1 \le i, j \le d}$ as the coordinates of $v_i(q')$ and $w_i(q')$, $1 \le i \le d$, in the canonical basis $(\delta_j)_{1 \le j \le d}$. That is,

$$ v_i(q') = \sum_{j=1}^{d} a_{ij}(q') \delta_j, \quad w_i(q') = \sum_{j=1}^{d} b_{ij}(q') \delta_j \quad \forall i \in \{1, \dots, d\}. $$

We denote by $A(q') = (A_{ij}(q'))_{1 \le i, j \le d}$ and $B(q') = (B_{ij}(q'))_{1 \le i, j \le d}$. Then, by the definition of dual bases, we have

$$ \forall (i, j) \in \{1, \dots, d\}^2 \quad (v_i(q'), w_j(q')) = \sum_{k=1}^{d} a_{ik}(q') b_{jk}(q') = \begin{cases} 1 & \text{if } i = j, \\ 0 & \text{otherwise,} \end{cases} $$
and thus $A(q')B(q') = \text{Id}_{\mathbb{R}^d}$. We infer that $B(q') = A^{-1}(q')$. But, the mapping

$$ \mathcal{I} : \begin{cases} GL(\mathbb{R}^d) \to GL(\mathbb{R}^d) \\ Q \mapsto Q^{-1} \end{cases} $$

is continuous, and the mapping $q' \mapsto A(q')$ is continuous on $\bar{B}(q, r_q)$ with values in $GL(\mathbb{R}^d)$. It follows that $q' \mapsto B(q')$ is also continuous on $\bar{B}(q, r_q)$ and we infer that the mappings $w_j, j \in \{1, \dots, d\}$, (which are the columns of $B$) are also continuous on $\bar{B}(q, r_q)$.

It follows that we can define

$$ C_{*,q} = \max \{|w_j(q')|; q' \in \bar{B}(q, r_q)\}. $$

Now, using the compactness of $K \cap \mathcal{B}$, we infer that there exists a finite set of points $(q_k)_{1 \le k \le \ell}$ such that $q_k \in K \cap \mathcal{B}$ for all $k \in \{1, \dots, \ell\}$ and

$$ K \cap \mathcal{B} \subset \bigcup_{k=1}^{\ell} B(q_k, r_{q_k}). $$

Then, the conclusion follows with $C_{*,\mathcal{B}} = \max_{1 \le k \le \ell} C_{*,q_k}$. $\square$

**Lemma 8.** Let us recall the definition of $T_K(q)$:

$$ T_K(q) = \{w \in \mathbb{R}^d; (\nabla f_\alpha(q), w) \ge 0 \ \forall \alpha \in J(q)\} \quad \forall q \in \mathbb{R}^d $$

with

$$ J(q) = \{\alpha \in \{1, \dots, \nu\}; f_\alpha(q) \le 0\}. $$

Then for all $q_0 \in K$, there exist $\delta > 0$, $r > 0$ and $a \in \mathbb{R}^d$ such that, for all $q \in \bar{B}(q_0, 2\delta)$:

$$ \bar{B}(a, r) \subset T_K(q). \qquad (53) $$

**Proof.** Let $q_0$ be in $K$.

Since the functions $(f_\alpha)_{\alpha=1,\dots,\nu}$ are continuous, we infer that there exists $\delta_1 > 0$ such that, for all $\alpha \notin J(q_0)$, we have

$$ f_{\alpha}(q) > 0 \quad \text{if} \quad |q - q_0| \le \delta_1. $$

It follows that $J(q) \subset J(q_0)$ for all $q \in \bar{B}(q_0, \delta_1)$.

Consequently, if $J(q_0) = \emptyset$, we have $J(q) = \emptyset$ for all $q \in \bar{B}(q_0, \delta_1)$ and (53) is satisfied for $\delta = \delta_1/2$ and for all $a \in \mathbb{R}^d$ and $r > 0$.

Let us assume now that $J(q_0) \ne \emptyset$. For all $\alpha \in J(q_0)$ we define $\phi_\alpha : \mathbb{R}^d \times \mathbb{R}^d \to \mathbb{R}$ by

$$ \phi_{\alpha}(q, y) = (\nabla f_{\alpha}(q), y) \quad \forall (q, y) \in \mathbb{R}^{d} \times \mathbb{R}^{d} $$

and $\phi : \mathbb{R}^d \times \mathbb{R}^d \to \mathbb{R}$ by

$$ \phi(q, y) = \min_{\alpha \in J(q_0)} \phi_{\alpha}(q, y) \quad \forall (q, y) \in \mathbb{R}^d \times \mathbb{R}^d. $$
Since $f_\alpha \in C^1(\mathbb{R}^d)$ for all $\alpha \in \{1, \dots, v\}$, we obtain that the mappings are continuous. Moreover, since $(\nabla f_\alpha(q_0))_{\alpha \in J(q_0)}$ is linearly independent, we can define a basis $(\xi_i)_{1 \le i \le d}$ of $\mathbb{R}^d$ such that

$$ \xi_{\alpha} = \nabla f_{\alpha}(q_0) \quad \forall \alpha \in J(q_0). $$

Let us denote by $(\zeta_i)_{1 \le i \le d}$ the dual basis of $(\xi_i)_{1 \le i \le d}$ and let

$$ a = \sum_{\alpha \in J(q_0)} \zeta_{\alpha}. $$

Then, for all $\alpha \in J(q_0)$, we have

$$ \phi_{\alpha}(q_0, a) = (\nabla f_{\alpha}(q_0), a) = \left( \xi_{\alpha}, \sum_{\beta \in J(q_0)} \zeta_{\beta} \right) = 1 $$

and $\phi(q_0, a) = 1$. By continuity, it follows that there exist $r > 0$ and $\delta_2 > 0$ such that

$$ \phi(q, y) > 0 \quad \forall (q, y) \in \bar{B}(q_0, \delta_2) \times \bar{B}(a, r). $$

Let $\delta = \frac{1}{2} \min(\delta_1, \delta_2)$. For all $q \in \bar{B}(q_0, 2\delta)$ we have

$$ J(q) \subset J(q_0), \quad \phi(q, y) = \min_{\alpha \in J(q_0)} (\nabla f_{\alpha}(q), y) > 0 \quad \forall y \in \bar{B}(a, r) $$

which implies that

$$ \bar{B}(a, r) \subset T_K(q) = \{ y \in \mathbb{R}^d ; (\nabla f_\alpha(q), y) \ge 0 \quad \forall \alpha \in J(q) \} $$

and (53) is satisfied. $\square$

*Acknowledgements.* The author would like to thank the referee for the numerous relevant remarks which have contributed to improving the final form of the manuscript.

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