# On the numerical approximation of a vibration problem with unilateral constraints

Michelle Schatzman, Michel Bercovier

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## To cite this version:

Michelle Schatzman, Michel Bercovier. On the numerical approximation of a vibration problem with unilateral constraints. [Research Report] Ecole Polytechnique (Palaiseau, France). 1985. hal-03264087

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Distributed under a Creative Commons Attribution 4.0 International License
ON THE NUMERICAL APPROXIMATION OF A VIBRATION PROBLEM
WITH UNILATERAL CONSTRAINTS

«« »»

Michella SCHATZMAN *

Michel BERCOVIER **

(Février 1985)

<footnote n="1">¹ Centre de Mathématiques Appliquées, Ecole Polytechnique, 91128 Palaiseau Cedex - France.
present address : Mathématiques, Université Lyon 1 - Claude-Bernard, 69622 Villeurbanne Cedex,
FRANCE</footnote>

<footnote>** School for Applied Science. The Hebrew University of Jerusalem - Israel.</footnote>
# I - INTRODUCTION.

Consider the following problem : a rod of length L at rest vibrates longitudinally. One end of the rod is fixed, and the other one is free to move, as long as it does not hit a material obstacle, which may constrain the displacement of this end either to be greater than or equal, or to be lesser than or equal to some given number.

<figure><img src="image_3.png" /><figcaption>Figure 1.</figcaption></figure>

Let $x$ be the spatial coordinate along the rod, with the origin at the fixed end, let $u(x, t)$ be the displacement at time $t$ of the material point of spatial coordinate (at rest) $x$, and let $f$ denote a density of exterior forces, depending on space and time.

We can model this situation as follows : assume that the material of the rod is elastic, homogeneous, linear, and use the approximation of small displacements ; them, after scaling, so that the velocity of waves in the rod will be one, the following equation is satisfied

$$ (1.1) \qquad \Box u = u_{tt} - u_{xx} = f \quad \text{in } Q_T = (0, L) \times (0, T). $$

The boundary condition are determined as follows : at the fixed end, we have a Dirichlet condition

$$ (1.2) \qquad u(0, t) = 0; $$

assume an inferior constraint, for instance, at the other end, then

$$ (1.3) \qquad u(L, t) \geq k_0. $$
When the rod touches the obstacle, its reaction can be only upwards (see fig. 1), so that

$$0 \quad u_x(L,t) \ge 0 \quad \text{on} \quad \{t/u(L,t) = k_0\};$$

when the rod does not touch the obstacle, the end is free so that

$$u_x(L,t) = 0 \quad \text{on} \quad \{t/u(L,t) > k_0\}.$$

These last two conditions, which will be entirely justified in the sequel, can be summarized as

$$(1.4) \qquad u_x(L,t) \ge 0;$$

$$(1.5) \qquad u_x(L,t) (u(L,t) - k_0) = 0.$$

Conditions (1.3) - (1.5) are usually termed "unilateral conditions".

We are given initial conditions

$$(1.6) \qquad u(x,0) = u_0(x)$$

$$(1.7) \qquad u_t(x,0) = u_x(x)$$

(1.9)

It has been shown in theorem 14 of [5] that, if $u$ belongs to the Sobolev space $H^{3/2}(0, L)$, with $u_0(0) = 0$, $u_0(L) \ge k_0$, if $u_t$ belongs to $H^{1/2}(0, L)$ and $f$ belongs to $H^{3/2}(Q_T)$, then (1.1) - (1.7) possesses a unique solution $u$ in the space

$$\{u \in L^\infty(0, T; H^{3/2}(0, L)) / u \in L^\infty(0, T; H^{1/2}(0, L))\};$$

moreover the energy of this solution is conserved:

$$(1.8) \qquad \left\{
\begin{aligned}
& \frac{1}{2} \int_0^L \left( |u_t(x,t)|^2 + |u_x(x,t)|^2 \right) dx = \\
& \frac{1}{2} \int_0^L \left( |u_{x_1}(x)|^2 + \left| \frac{du}{dx}(x) \right|^2 \right) dx + \int_0^T \int_0^L u_t(x,s) f(x,s) dx ds, \\
& \forall t \in [0, T].
\end{aligned}
\right.$$
Observe that the model considered here does not include the possibility of a loss of energy during the contact with the obstacle, as (1.8) is a consequence of (1.1)-(1.7).

We shall show in this paper that, firstly $u_0, u_1$ and $f$ can be taken in larger functional spaces than the ones mentioned above, and secondly that the solution $u$ of (1.1)-(1.7) is the solution of a variational inequality. Though an explicit solution of (1.1)-(1.7) can be written (see [5]) because of the simple geometry of the problem, the variational formulation enables us to replace if necessary the homogeneous rod by an inhomogeneous one, and to build in a very natural fashion numerical schemes, which eventually will be shown to converge, with the help of very classical tools.

In section 2 of this paper, we given an equivalent variational formulation of (1.1)-(1.7), which uses several functional lemmas, and we prove the existence of a solution by Galerkin's method of approximation in finite dimensional space in the beginning of section 3.

In the remaining part of section 3, we consider a family of finite element schemes in a variational formulation written as

$$ u^{n+1} \in K_h = K \cap V_h $$

$$ (1.9) \qquad \left( \frac{u^{n+1} - 2u^n + u^{n-1}}{\Delta t^2}, v - u^{n+1} \right) + a(u^n, v - u^{n+1}) \ge (f^n, v - u^{n+1}), \forall v \in V_h, $$

where $K$ is the convex of constraints, $V_h$ is the space of finite clements, and $a$ is the bilinear form

$$ a(u, v) = \int_0^L u_x v_x dx; $$

We show that under a suitable stability condition, the approximation defined by (1.9) converges.

In section 4, we turn to explicit finite difference schemes, which can be written in variational form, and are analogous to (1.9), with the main difference that the mass matrix is diagonal, so that the numerical implementation is extremely easy: these schemes are given by:
$$ \frac{u_{j}^{k+1} - 2u_{j}^{k} + u_{j}^{k-1}}{\Delta x^{2}} - \frac{u_{j+1}^{k} - 2u_{j}^{k} + u_{j}^{k-1}}{\Delta t^{2}} = F_{j}^{k} \quad \text{if } j < n $$

$$ u_{n}^{k+1} = \max \left[ 2 u_{n}^{k} - u_{n}^{k-1} + \frac{\Delta t^2}{\Delta x^2} (u_{n-1}^{k} - u_{n}^{k}) + \Delta t^2 F_{j}^{k}, k=0 \right] $$

The proof of convergence is fairly easy if the scheme is non characteristic ($\Delta t < \Delta x$); it is much more technical in the characteristic case ($\Delta t = \Delta x$).

In section 5, we report on numerical experiments. They can be summarized as follows: if the initial data are smooth, the characteristic and the non-characteristic explicit schemes give reasonably good results; the characteristic scheme is substantially better; this suggests that for non-homogeneous rods, the CFL number should be taken as large as possible. If the initial data are piecewise affine, the characteristic scheme gives a good result, and the noncharacteristic scheme, which should have a regularizing effect gives an absolutely terrible result. This suggests that the computation of the phenomena with rough data, in a non-homogeneous medium, will be difficult.

Before concluding this introduction, let us show how Problem (1.1)-(1.8) is closely related to the problem of the vibrating string with pointwise unilateral constraints which has been studied in [1], [3], [4], [10].

Let $u$ be a solution of (1.1)-(1.7) and denote by $\tilde{u}$ the function defined by

$$ \tilde{u}(x,t) = u(x,t), \quad 0 \le x \le L $$

$$ \tilde{u}(x,t) = u(2L - x, t), \quad L \le x \le 2L. $$

and the initial conditions $u_0$ and $u_1$ are similarly symmetrized.

Then, it is easy to check that $\tilde{u}$ satisfies the following relations

$$ (1.10) \qquad \tilde{\mu} = \mu > 0 \quad (\mu \text{ is a measure}) $$

$$ (1.11) \qquad \tilde{u}(L,t) > k_0 $$

$$ (1.12) \qquad \operatorname{supp}(\mu) \subset \{L\} \times \{t \in [0, T] / \tilde{u}(L,t) = k_0\} $$
$$
(1.13) \qquad \tilde{u}(x, 0) = \tilde{u}_0(x)
$$

$$
(1.14) \qquad \tilde{u}_t(x, 0) = \tilde{u}_1(x),
$$

$$
(1.15) \qquad \tilde{u}(0,t) = \tilde{u}(2L,t) = 0 .
$$

These are exactly the equations of a vibrating string with a peg of height $k_0$ at $x = \frac{L}{2}$;
in [10] where this problem, and more general ones were studied, a rather awkward variational formulation was given. It is our hope that the one which is given here will prove more useful.
## II - THE VARIATIONAL FORMULATION.

### II.1 - NOTATIONS AND DEFINITIONS.

We shall need the following spaces and sets:

$$ (2.1). \qquad V = \{ u \in H'(0, L) / u(0) = 0 \}, $$

with scalar product defined by

$$ (2.2) \qquad ((u, v)) = \int_{0}^{L} u' v' \, dx $$

and norm $|| \cdot ||$.

Let

$$ (2.3) \qquad H = L^2(0, L) $$

with scalar product $(\cdot, \cdot)$ and norm $| \cdot |$, and define a convex subset of $V$ by

$$ (2.4) \qquad K = \{ u \in V / u(L) \ge k_0 \}. $$

We define now spaces involving the time variable:

$$ (2.5) \qquad \mathcal{X}_2 = \{ u \in L^2(0, T, V) / u_t \in L^2(0, T, H) \} $$

with norm

$$ (2.6) \qquad \| u \|_2 = \left[ \int_0^T (\| u(t) \|^2 + |u_t(t)|^2) dt \right]^{1/2} $$

and

$$ (2.7) \qquad \mathcal{X}_\infty = \{ u \in L^\infty(0, T; V) / u \in L^\infty(0, T; H) \} $$

with norm

$$ \| u \|_{\infty} = \sup_{t \in [0, T]} \{ \| u(t) \| + |u_t(t)| / u \} \quad \text{c} \in [0, T] $$
Observe, that the space $\mathcal{X}_2$ is the space of functions of locally bounded energy; it topology is the topology of $H^1(Q_T)$, and

$$ (2.9) \qquad K = \{ u \in \mathcal{X}_2 / u(L, \cdot, t) \ge k \text{ a.e. } \} $$

makes sense; we have the identity

$$ K = \{ u \in \mathcal{X}_2 / u(\cdot, \cdot, t) \in K \text{ for almost every } t \}. $$

## II.2 - FUNCTIONAL RESULTS.

We need now two functional results which allow one to see first that $\mathcal{X}_\infty$ is a space of Hölder continuous functions and secondly that, if $u$ belongs to $\mathcal{X}_2$ and $\square u$ is square integrable, then it is possible to take traces of $u$ on $|t=0 \cup \{t=T\}$ and similarly $u_x$ on $|x=0 \cup \{x=L\}|$.

**Lemma II.1** - $\mathcal{X}_\infty$ is included in the space of Hölder continuous functions $C^{0,1/2}(Q_T)$ and the injection from $\mathcal{X}_\infty$ to $C^{0, 1/2}(Q_T)$ is continuous.

**Proof** - Let us prove first that any function of $\mathcal{X}_\infty$ is a.e. equal to a continuous function. After an eventual modification of $u \in \mathcal{X}_\infty$ on a null subset of $Q_T$,

$$ (2.10) \qquad u \in C^0([0, T], L^2(0, L)). $$

Let

$$ E = \{ t \in [0, T] / \| u(\cdot, t) \| \le \| u \|_\infty \} $$

The complement of $E$ in $[0, T]$ is null; therefore, for any $t$ in $[0, T]$, there exists a sequence $t_n$ in $E$ such that $t_n \to t$; possibly after extracting a new subsequence, there exists a $v$ in $V$ such that

$$ u(\cdot, t) \to v \text{ in } V \text{ weak.} $$
By comparison with (2.10)

$$u(\cdot, t) = v,$$

so that

$$u \subset C^0([0, T]; \text{V weak}).$$

For any $\epsilon$ in $(0, \frac{1}{2})$, the injection $V \subset H^{1-\epsilon}(0, L)$ is compact and $H^{1-\epsilon}(0, L)$ is a space of continuous functions on $[0, L]$. This proves our first claim.

To estimate the Holder norm of $u$, observe first that

$$|u(x,t) - u(x',t')| \le |\int_x^{x'} u(x,t') dy| < \sqrt{|x' - x|} \|u\|_{\infty}$$

Let $h < L/2$, and $x \in [0, L/2]$; we estimate the modulus of continuity in time by splitting $u(x,t_1) - u(x,t_2)$ in three pieces: the two outer ones involve a difference between $u(x,t_1)$ and a spatial average and the middle one a difference between spatial averages at times $t_1$ and $t_2$:

$$|u(x, t_i) - \frac{1}{h} \int_x^{x+h} u(x', t_i) dx' | \le \frac{1}{h} \int_x^{x+h} |u(x, t_i) - u(x', t_i)| dx' \le \sqrt{h} \|u\|_{\infty}$$

On the other hand

$$\begin{aligned} \left| \frac{1}{h} \int_x^{x+h} \left( u(x', t_1) - u(x', t_2) \right) dx' \right| &\le \frac{1}{h} \left| \int_x^{x+h} \int_{t_1}^{t_2} u_t(x', s) ds dx' \right| \\ &\le \|u\|_{\infty} |t_2 - t_1| h^{-\frac{1}{2}}. \end{aligned}$$

Therefore

$$|u(x, t_2) - u(x, t_1)| \le \|u\|_{\infty} \left( 2\sqrt{h} + |t_2 - t_1|/\sqrt{h} \right).$$
If we choose $h = |t_2 - t_1|^{1/2}$ and $|t_2 - t_1| \le \sqrt{L/2}$, then

$$|u(x, t_1) - u(x, t_2)| \le 3 \|u\|_{\infty} \sqrt{|t_2 - t_1|}$$

Patching the relations obtained on strips of height $(L/2)^{1/2}$ and the relations obtained for $L/2 < x < L$, we obtain

$$ (2.11) \qquad \left\{ \begin{array}{l} |u(x,t) - u(y,s)| \le K \|u\|_{\infty} \sqrt{|t-s| + |x-y|} \\ \forall (x,t) \in Q_T, (y,s) \in Q_T. \end{array} \right. $$

If we let $x = t = 0$ in (2.11), and integrate in $(y, t)$ we obtain the estimate

$$ (2.12) \qquad |u(0,0)| \le K' \|u\|_{\infty}; $$

relations (2.11) and (2.12) allow to conclude that there exists a constant $C$ such that

$$ \begin{align*} \max_{Q_T} |u(x,t)| + \max_{\substack{Q_T \times Q_T \\ (x,t) \neq (y,s)}} \frac{|u(x,t) - u(y,s)|}{(|x-y| + |t-s|)^{1/2}} &= \\ &= \|u\|_{0,\frac{1}{2}} \le C \|u\|_{\infty}. \end{align*} $$

To prove the trace result we need two alternative solutions: the first one is to play with energy inequalities, and after a good deal of manipulations, obtain the required results. This procedure is rather awkward, and probably too particular. The second one, which we choose, consists in defining weak solutions of evolution equations in the setting of the theory of strongly continuous semi-groups and their duals, and then proving a regularity result. This technique was chosen, because the regularity result is, in itself, interesting.

Let therefore $X$ be a reflexive Banach space with norm $\| \cdot \|$ and dual $X^*$. Let $A$ be a not necessarily bounded operator in $X$ with domain $D(A)$. If $A$ is the generator of a strongly continuous semigroup $S(t)$, there exist constants $M$ and $\omega$ such that
$$
(2.13) \qquad \left\{
\begin{array}{l}
\| S(t) \| \le M e^{\lambda t} , \quad \forall t \ge 0 \\
\| (\lambda + \lambda^{-n})^{-1} \| \le M (\lambda - \omega)^{-n} , \quad \forall \lambda \ge \omega, \forall n \ge 1
\end{array}
\right.
$$

We know that, for $u_0$ in $D(A)$, there exists a unique function $u$ in
$C^1([0, T]; X) \cap C^0([0, T]; D(A))$ such that

$$
(2.14) \qquad \frac{du}{dt}(t) + Au(t) = 0
$$

with initial condition

$$
(2.15) \qquad u(0) = u_0
$$

Then

$$
u(t) = S(t) u_0 .
$$

A function $u$ in $C^1([0, T]; X) \cap C^0([0, T]; D(A))$ which satisfies (2.14) is called a strong solution.

We know from [7] that, if $S^*(t)$ is defined by

$$
\langle S(t) u, u^* \rangle = \langle u, S^*(t) u^* \rangle, \quad \forall u \in X, \forall u^* \in X^*,
$$

then $S^*(t)$ is a strongly continuous semi-group, whose generator $A^*$ is given by

$$
D(A^*) = \{ u^* \in X^* / u \to <Au, u^*> \text{ is continuous on } D(A) \text{ in the norm topology of } X \}.
$$

is continuous on $D(A)$ in the norm topology of $X$.

and

$$
\langle u, A^* u^* \rangle = \langle Au, u^* \rangle, \quad \forall u \in D(A), \forall u^* \in D(A^*)
$$
We define a notion of weak solution as follows :

**Définition 1.2** - Let $u$ belong to $L_{\text{loc}}^1((0, T); X)$ and $f$ to $L^1(\{0, T\}; X)$. Then $u$ is said to be a weak solution of (2.14) if

$$ (2.17) \quad \left\{ \begin{array}{l} - \int_0^T \dot{\psi}(s) \langle u(s), u^* \rangle ds + \int_0^T \psi(s) \langle u(s), A^*u^* \rangle ds = \\ \\ = \int_0^T \psi(s) \langle f(s), u^* \rangle ds, \quad \forall \psi \in \mathcal{D}((0, T)), \forall u^* \in D(A^*) \end{array} \right. $$

Then we have the following regularity result on weak solutions :

**Proposition 1.3** - Let $u$ be a weak solution of (2.14) according to definition 1.2. Then, possibly after modifying $u$ on a null subset of $[0, T]$,

$$ u \in C^0([0, T]; X). $$

**Proof** - The function defined by

$$ v(t) := \int_0^t S(t-s) f(s) ds $$

is a weak solution of (2.15); to see this, let $\psi$ belong to $\mathcal{D}((0, T))$ and $u^*$ to $D(A^*)$; then

$$ - \int_0^T \psi(t) \langle v(t), u^* \rangle dt = - \int_0^T \psi(t) \int_0^t S(t-s) f(s), u^* ds dt, $$

and, by definition of $S^*$, this expression is equal to

$$ - \int_0^T \psi(t) \int_0^t \langle f(s), S^*(t-s) u^* \rangle ds dt = \int_0^T \langle f(s), \int_S^T S^*(t-s) u^* \psi(t) ds \rangle ds. $$
But, as $u^*$ belongs to $D(A^*)$, we may integrate by parts

$$ - \int_s^T S^*(t-s) u^* \psi(t) dt = \psi(s) u^* - \int_s^T A^* S^*(t-s) u^* \psi(t) dt $$

Therefore,

$$ \begin{align*} & - \int_0^T \psi(t) \langle v(t), u^* \rangle dt + \int_0^T \psi(t) \langle v(t), A^* u^* \rangle dt \\ &= \int_0^T \psi(s) \langle f(s), u^* \rangle ds - \int_0^T \langle f(s), \int_0^T A^* S^*(t-s) u^* \rangle \psi(t) dt ds \\ &\quad + \int_0^T \psi(t) \langle \int_0^t S(t-s) f(s), A^* u^* \rangle ds dt, \end{align*} $$

which proves by exchanging the order of integration in the last term that $v$ is a weak solution.

Clearly, $v$ belongs to $C^0([0, T]; X)$, and $u - v$ is a weak solution of (2.14), with zero right hand side. Therefore, it is enough to prove that weak solutions of (2.15) with zero right hand side are continuous from $[\varepsilon, T]$ to $X$, for all positive $\varepsilon$. Let $\lambda > \omega$, and take in (2.17), with the help of (2.13),

$$ v^* \in D(A) \quad , \quad u^* = (A^* + \lambda)^{-2} v^* . $$

Clearly, $u^*$ belongs to $D((A^*)^3) \subset D(A^*)$, and we may check immediately that, for any $u$ in $X$

$$ (2.18) \qquad \langle \bar{u}, (A^* + \lambda)^{-2} v^* \rangle = \langle (A + \lambda)^{-2} \bar{u}, v^* \rangle $$

$$ (2.19) \qquad \langle \bar{u}, A^*(A^* + \lambda)^{-2} v^* \rangle = \langle A(A + \lambda)^{-2} \bar{u}, v^* \rangle . $$

Therefore, if we let

$$ (2.20) \qquad z(t) = (A + \lambda)^{-2} u(t) $$

$z$ is a weak solution of (2.15), and we can see that

$$ (2.21) \qquad \frac{dz}{dt} + Az = 0 \quad \text{in the sense of distributions.} $$
By definition of $z$, if $D(A^n)$ is equipped with the norm
$$ \sum_{j=0}^{n} \| A^j x \| $$

$$ (2.22) \qquad z \in L_{\text{loc}}^{1}((0, T); D(A^2)) $$

so that, from (2.21),

$$ \frac{dz}{dt} \in L_{\text{loc}}^{1}((0, T); D(A)), $$

Therefore, after an eventual modification of $z$ on a null subset of $[0, T]$,

$$ z \in C^0((0, T); D(A)). $$

Still with (2.21),

$$ \frac{dz}{dt} \in C^0((0, T); X) $$

which implies that

$$ z \in C^1((0, T); X) $$

Property (2.16) for a strong solution implies that

$$ (2.23) \qquad z(t) = S(t-s)z(s) \quad \forall s>0, \forall t>s. $$

Pick an $s > 0$ such that

$$ \| A^2 z(s) \| < +\infty. $$

Then

$$ S(t-s)z(s) \in D(A^2), \forall t \ge s $$

and

$$ A^2[S(t-s)z(s) - z(s)] = S(t-s)A^2z(s) - A^2z(s), $$

and by the strong continuity of $S$,
$$z \in C^0([s, T]; D(\Lambda^2)).$$

As the set where $\| \Lambda^2 z(s) \|$ is infinite is null, we obtain

$$z \in C^0([\epsilon, T]); X), \forall \epsilon > 0,$$

and the conclusion of the Proposition follows. ■

The second functional lemma is the following trace lemma:

**Lemma 1.4** - Let $u$ belong to $X_2$, and assume that $\Box u = u_{tt} - u_{xx}$ (defined in the sense of distributions on $Q_T$) is square integrable.

Then, for every positive $\epsilon$,

$$(2.24) \quad u \in C^0([0, T]; H^1(0, L - \epsilon)) \text{ and } u_t \in C^0([0, T]; L^2(0, L - \epsilon))$$

$$(2.25) \quad u \in C^0([0, L]; H^1(\epsilon, T - \epsilon)) \text{ and } u_x \in C^0([0, L]; L^2(\epsilon, T - \epsilon))$$

**Proof** - Let $\psi$ belonging to $\mathcal{D}([0, L])$ be such that

$$\psi = 1 \text{ on } [0, L - \epsilon]$$

$$\psi = 0 \text{ on } [L - \epsilon/2, L]$$

Let

$$\tilde{u} = u\psi.$$

Then

$$\Box \tilde{u} = \tilde{f} \in L^2(Q_T)$$

$$\tilde{u}(0, T) = \tilde{u}(L, t) = 0.$$

If we let

$$X = H_0^1(0, L) \times L^2(0, L)$$
and

$$D(\mathcal{A}) = [H_0^1(0, L) \cap H_0^2(0, L)] \times H_0^1(0, L)$$

with

$$\mathcal{A} \begin{pmatrix} u \\ v \end{pmatrix} = \begin{pmatrix} -v \\ u_{xx} \end{pmatrix} \quad \forall \mathcal{U} = \begin{pmatrix} u \\ v \end{pmatrix} \in D(\mathcal{A}).$$

it can be checked immediately that $\mathcal{A}$ is skew-adjoint, and therefore, thanks to Stone's theorem, $\mathcal{A}$ generates a group of isometries $\mathcal{S}(t)$ on $X$. If we let

$$\mathcal{U} = \begin{pmatrix} \tilde{u} \\ \tilde{u}_t \end{pmatrix}, \quad \mathcal{F} = \begin{pmatrix} 0 \\ \tilde{f} \end{pmatrix}$$

then, $\mathcal{U}$ is a weak solution of $\partial_t\mathcal{U}/\partial t + \mathcal{A}\mathcal{U} = \mathcal{F}$, and, according to proposition (2.3), and using the group property of $\mathcal{S}$, $\mathcal{U}$ belongs to $C^0([0, T]; X)$, which proves (2.24).

To prove (2.25), we take advantage of the symmetric form of the wave equation with respect to $x$ and $t$ in one spatial dimension, and we multiply $\mathcal{U}$ by some $\chi$ in $\mathcal{D}((0, T))$ such that $\chi = 1$ in $[\varepsilon, T - \varepsilon]$, and $\chi = 0$ in $[0, \varepsilon/2] \cup [L - \varepsilon/L, L]$; then, we argue exactly as above. ■

## II.3 - THE EQUIVALENCE BETWEEN THE ORIGINAL FORMULATION OF THE PROBLEM AND THE VARIATIONNAL FORMULATION.

This equivalence is summarized in the following statement :

**Theorem 2.5** - Let $f$ belong to $L^2(Q_T)$, $u_0$ to $K$ and $u_1$ to $H$. Let $u$ belong to $X$; then $u$ is a solution of (1.1) - (1.7) iff

$$(2.26) \qquad u(0) = u_0$$

$$(2.27) \qquad u \in K$$
$$
(2.28) \quad \left\{
\begin{array}{l}
-(u_1, v(0) - u_0) - \int_0^T (u_t, v_t - u_t) dt + \int_0^T ((u, v - u)) dt \ge \\
\ge \int_0^T (f, v - u) dt, \\
\forall v \in K \text{ such that there exists } \eta > 0 \text{ such that } v = u \text{ for } t \ge T - \eta
\end{array}
\right.
$$

Proof - We check first that (2.26) - (2.28) make sense :

As $u$ and $v$ belong to $\mathcal{X}_2$, they have traces in $L^2(0, L)$ on $\{t=0\} \times (0, L)$, so
that (2.26) and the first term of (2.28) make sense. The remaining terms of inequality
make sense quite immediately.

Assume now that $u$ and $f$ satisfy the hypotheses of the Proposition and relations (2.26) and (2.28). Then, if we pick any $\varphi$ in $\mathcal{D}(Q_T)$, $u + \varphi = v$ can be employed as a test function in (2.28), so that

$$
(2.29) \qquad u_{tt} - u_{xx} = f \quad \text{in the sense of distributions in } Q_T
$$

From Lemma 1.4, the trace of $u$ on $\{t=0\}$ exists and belongs to $L^2_{\text{loc}}((0, L))$
and the trace of $u_x$ on $\{x=L\}$ exists and belongs to $L^2_{\text{loc}}((0, T))$. Let $\eta$ be fixed,
and let

$$
\mathcal{X}_2^\epsilon = \{ \varphi \subset \mathcal{X}_2 / \varphi = 0 \text{ for } t \ge T - \eta \text{ or}
$$

$$
\text{for } x \le \epsilon \quad \text{or for} \quad L - x + t \le \epsilon
$$

The definition of $\bar{\mathcal{X}}_2^\epsilon$ is precisely tailored to avoid any trouble with the (possible) blow-up of $u_x$ and $u_t$ in the corner $(L, 0)$ of $Q_T$.

If $\varphi$ belong to $C^\infty(\overline{Q}_T)$, $\cap \mathcal{X}_2^\epsilon$, the following identities hold by density :

$$
(2.30) \quad \langle u_t, \varphi \rangle = -(u_t(0), \varphi(0)) - \int_0^T (u_t, \varphi_t) dt
$$
$$ (2.31) \qquad \langle u_{xx}, \varphi \rangle = \int_0^T u_x(L,t) \varphi(L,t) dt - \int_0^T ((u, \varphi)) dt. $$

Still by a density argument, the above relations hold for any $\varphi$ in $\chi_2^\epsilon$. If we use relations (2.30) and (2.31) in (2.28) with the test function $v = u + \varphi$, $\varphi \in \chi_2^\epsilon$, $u + \varphi \in K$, we obtain.

$$ (2.32) \qquad (u_t - u_{tt}(0), v(0) - u_0) \le \int_0^T u_x(L,t)(v(L,t) - u(L,t)) dt. $$

Choose first $\varphi$ vanishing on $x = L$; then we deduce from (2.32) that

$$ u_t(0) = u_x(0), \quad \text{a.e. on } (0, L), $$

and, in particular, the trace of $u_t$ on $\{t=0\}$ is square integrable. Now (2.32) becomes

$$ (2.33) \qquad \begin{cases} \int_0^T u_x(L,t)(v(L,t) - u(L,t)) dt \ge 0 \\ \forall v \in K \quad \text{such that } v - u \in \chi_2^\epsilon. \end{cases} $$

According to Lemma 1.4, $u$ has a trace on $\{x = L\}$, which belongs to $H_{\text{loc}}^1((0, T))$ and $u_x$ has a trace on $\{x = L\}$ which belongs to $L_{\text{loc}}^2((0, L))$. If we define $v$ by

$$ v = u + \psi k $$

where

$$ \psi(x) = \begin{cases} 1 & \text{for } x \ge L - \epsilon \\ 0 & \text{for } x < L - 2\epsilon \end{cases} \\ \psi \in \mathcal{H}([0, L]) $$

and

$$ k \in H^1((0, T)), k(x) = 0 \text{ for } x \le \epsilon \text{ or } x \ge T - \epsilon, $$
then $v$ is a convenient test function in (2.33), and a classical argument [6] shows that

$$ u_x(L,t) \ge 0 \quad \text{a.c. on } [0,T] $$

$$ u(L,t) \ge 0 \quad \forall t \in [0,T] $$

$$ u_x(L,t) (u(L,t) - k) = 0 \quad \text{a.e. on } [0,T] $$

i.e. (1.3) - (1.5).

Let us now prove the converse: assume that $u$ satisfies (1.1) – (1.7). As $u$ belongs to $X_2$ and $f$ to $L^2(Q_T)$, $u$ and $u_x$ have a trace on $\{x=L\}$ so that the unilateral conditions (1.3) – (1.5) make sense; similarly, as $u$ and $u_x$ have a trace on $\{t=0\}$, the initial conditions (1.6) and (1.7) make sense. Relations (2.26) and (2.27) are quite clear, and one has only to check (2.28). The difficulty comes from the fact that, a priori, $t \to u_x(L,t)$ is only locally in $L^2$, so that we shall have to prove that $u_x(L,\cdot)$ is in $L^2$ indeed. We have even a more precise statement:

**Lemma 1.6** - Let $u$ satisfy (1.1) – (1.7). Then, for all positive $\varepsilon$, $u \in C^0([0,L]; H^1(0,T-\varepsilon))$, $u_x \in C^0([0,L]; L^2(0,T-\varepsilon))$.

*Proof* - We have the following energy identity, proved by a density argument:

$$ (2.36) \qquad (u_x^2 + u_t^2) - 2(u_x u_t - u_t u_x) = -2 f u_x \quad \text{in the sense of distributions.} $$

We may integrate it on $[x_0, x_1] \times [t_0, t_1]$ for $0 < x_0, x_1$ and $0 < t_0 < t_1$, thanks to (2.24) and (2.25), and we obtain

$$ (2.35) \quad \begin{cases} \int_0^t [(u_x^2 + u_t^2)(x_1, t) - (u_x^2 + u_t^2)(x_0, t)] dt = \\ = \int_{x_0}^{x_1} [(2u_x u_t)(x, t_1) - (2u_x u_t)(x, t_0)] - 2 \int_0^t f u dt] dx \end{cases} $$
The right hand side of (2.35) is bounded independently of $x_0, x_1, t_0, t_1$ as long as $x_0, x_1 \le \bar{x}_1 < L$, thanks to (2.24). If we integrate (2.35) with respect to $x$ on $[0, \bar{x}_1]$, we obtain

$$ \bar{x}_1 \int_0^t (u_x^2 + u_t^2) (x_1, t) \, dt \le \int_0^t \bar{x}_1 \int_0^t (u_x^2 + u_t^2) (x, t) \, dt + C \bar{x}_1, $$

which proves that

$$ x \to \int_0^t (u_x^2 + u_t^2) (x, t) \, dt \text{ is bounded on } [0, \bar{x}_1] $$

independently of $t_0$ and $t_1$.

and therefore,

$$ x \to \int_0^T (u_x^2 + u_t^2) (x, t) \, dt \text{ is bounded on } [0, \bar{x}_1]. $$

If we let $v$ be the solution of

$$ \square v = 0 \quad \text{on } (x_1, L) \times (0, T) $$

$$ \left. \begin{array}{l} v(x_1, t) = u(x_1, t) \\ v_x(x_1, t) = u_x(x_1, t) \end{array} \right\} t \in [0, T] $$

$$ \left. \begin{array}{l} v(x, 0) = u_0(x) \\ v(x, T) = u(x, T) \end{array} \right\} x \in [x_1, L] $$

Then

$w = u - v$ is the solution of
$$ \square w = f \quad \text{on } (x_1, L) \times (0, T) $$

$$ w(x_1, t) = w_x(x_1, t) = 0, \ t \in [0, T] $$

$$ w(x, 0) = w(x, T) = 0 \quad x \in [x_1, L] $$

and we may apply Proposition 2.3 to get

$$ w \in C^0([x_1, L]; H^1(0, T)) $$

$$ w \in C^0([x_1, L]; L^2(0, T)). $$

On the other hand, it is immediate that

$$ v \in C^0([x_1, L]; H^1(0, T + x_1 - L)) $$

$$ v_x \in C^0([x_1, L]; L^2(0, T + x_1 - L)). $$

Therefore, we obtain that, for all positive $\epsilon$,

$$ (2.36) \quad \left\{ \begin{array}{l} u \in C^0([0, L]; H^1(0, T - \epsilon)) \\ u_x \in C^0([0, L]; L^2(0, T - \epsilon)). \end{array} \right. $$

We resume now the proof of the converse of the Theorem. For $\varphi$ in $\chi_2^E$, we may apply identities (2.30) and (2.31) which imply

$$ (2.37) \quad \left\{ \begin{array}{l} -(u_t(0), \varphi(0)) - \int_0^T (u_t, \varphi_t) \, dt - \int_0^T u_x(L, T) \varphi(L, t) \, dt \\ \qquad + \int_0^T ((u, \varphi)) \, dt = \int_{Q_T} (f, \varphi) \, dx. \end{array} \right. $$

Let $\varphi$ belong to $\chi_2$ such that $\varphi(x, t) = 0$ for $t \ge T - \eta$. Let us extend $\varphi$ as follows:
$$ \varphi(x, t) = -\varphi(-x, t) \quad \text{for } -\alpha_0 \le x \le 0 $$

$$ \varphi(x, t) = \varphi(2L - x, t) \quad \text{for } L \le x \le L + \alpha_0 $$

$$ \varphi(x, t) = 0 \quad \text{for } T \le t \le T + \alpha_0 $$

where $\alpha_0$ is some small positive number. Let $\rho$ be a smooth convolution kernel :

$$ \rho(x, t) = \rho_1(x) \rho_2(t) $$

with

$$ \text{supp } \rho_1 = [-\frac{1}{2}, \frac{1}{2}], \quad \text{supp } \rho_2 = [-1, 0] $$

We convolve $\varphi$ with $\rho_\alpha(x, t) = \alpha^{-2} \rho(x \alpha^{-1}, t \alpha^{-1}) :$

$$ \varphi_{\alpha}(x, t) = \int \varphi(y, s) \rho_{\alpha}(x - y, t - s) dy ds $$

Then $\varphi_\alpha$ is well defined on $Q_T$ for $\alpha \le \alpha_0$.

Let $\chi$ be defined by

$$ \chi(x) = \int_{-\infty}^{x} \rho_{2}(-s) \, ds, $$

so that

$$ \chi(x) = 0 \quad \text{for } x \le 0 $$

$$ \chi(x) = 1 \quad \text{for } x \ge 1. $$

Then, we define

$$ \psi(x, t) = \chi(\epsilon^{-1}(x - \epsilon)) \chi(\epsilon^{-1}(t - x + L - \epsilon)) $$

We can see that $\varphi_{\alpha} \psi_{\epsilon} |_{Q_T}$ belongs to $\chi^{\epsilon}$, and (2.37) holds. Let us show that $\varphi_{\alpha} \psi_{\epsilon}$ converges to $\varphi_{\alpha}$ in the weak topology of $\chi_2$, as $\epsilon$ goes to zero.
$$ \frac{\partial}{\partial x} (\varphi_{\alpha} \psi_{\varepsilon}) = \frac{\partial \varphi_{\alpha}}{\partial x} \psi_{\varepsilon} + \varphi_{\alpha} \frac{\partial \varphi_{\varepsilon}}{\partial x}, $$

The first term of the right hand side converges strongly to $\frac{\partial \varphi_{\alpha}}{\partial x}$ in $L^2(Q_T)$, thanks to Lebesgue's theorem. Let us estimate the second term : thanks to the odd extension of $\varphi$ around $x=0$, we have

$$ \varphi_{\alpha}(0, t) = 0, $$

and thus, there exists a constant $C_1$ such that

$$ \| \varphi_{\alpha}(x, t) \| \le C_1 |x|. $$

Therefore,

$$ \begin{align*} \int_0^T \int_\varepsilon^{2\varepsilon} |\varphi_\alpha(x,t)| \frac{\partial \psi_\varepsilon}{\partial x}(x,t)^2 dx dt &\le \\ &\le T C_1^2 (\max \rho_2)^2 \varepsilon^{-2} \int_\varepsilon^{2\varepsilon} x^2 dx \\ &\le 3T C_1^2 (\max \rho_2)^2 \varepsilon. \end{align*} $$

Similarly

$$ \begin{align*} & \int_{0}^{T} \int_{\varepsilon}^{2\varepsilon} |\varphi_{\alpha}(x,t)| \frac{\partial \psi_{\varepsilon}}{\partial x}(x,t)^2 dx dt \le \\ & \qquad \varepsilon < t - x + L < 2\varepsilon \\ & \le C_1^2 L^2 (\max \rho_2)^2 \varepsilon^{-2} \frac{3\varepsilon^2}{2} \le \\ & \qquad 2 C_1^2 L^2 (\max \rho_2)^2. \end{align*} $$

This proves that $(\varphi_{\alpha}, \psi_{\varepsilon})$ is bounded in $L^2(Q_T)$ independently of $\varepsilon$; the same kind of arguments shows that $(\varphi_{\alpha}, \psi_{\varepsilon})$ is bounded in $L^2(Q_T)$ independently of $\varepsilon$. As $\varphi_{\alpha}, \psi_{\varepsilon}$ converges to $\varphi_{\alpha}$ in $D^1(Q_T)$, when $\varepsilon$ goes to zero, it must converge in the weak topology of
$H^1(Q_T)$ to $\phi$. On the other hand, $\phi_\alpha \psi_\epsilon$ converges to $\phi_\alpha$ in $L^2(0,T)$, and $(2.37)$ holds for $\phi = \phi_\alpha$. As $\alpha$ goes to zero, we may still pass to the limit because

$$ \phi_\alpha \to \phi \text{ in } H^1(Q) \text{ strong} $$

and the trace $\phi_\alpha(L, \cdot)$ converges to the trace $\phi(L, \cdot)$ thanks to the trace theorems in Sobolev spaces; observe that we do not have trouble with the corner $(L, 0)$ because the kernel $\rho$ is shifted towards the negative $t$, so that $\phi_\alpha(x, t)$ involves only points with $t' \ge t$; we do not have any more troubles with the corner $(L, T)$ because $\phi_\alpha$ vanishes identically for $t \ge T - \eta + \alpha$.

Then (2.37) holds for any $\phi$ in $X_2$. If we choose $v$ in $K$, then, (2.37) holds for $\phi = v - u$, and the term $\int_0^T u_x(L,t)(v(L,t) - u(L,t))dt$ is non-negative in virtue of (1.3)-(1.5), and this concludes the proof of theorem 2. ■

For the proof of uniqueness, the reader is referred to [5], where the uniqueness is a consequence of the formulation of our problem as a monotone problem on the boundary.

### III. THE EXISTENCE THEOREM; CONVERGENCE OF FULLY DISCRETIZED FINITE ELEMENTS METHODS.

#### III.1 - THE EXISTENCE THEOREM : APPROXIMATION BY FINITE DIMENSIONAL SPACES.

Let $(V_h)$ be a sequence of finite dimensional subspaces of $V$, of finite dimension, such that

$$ (3.1) \qquad \overline{U_v}_h^\nu = V_h. $$

Let $K$ be the sequence of closed convex subsets of $V_h$ given by

$$ (3.2) \qquad K_h = K \cap v_h. $$
We approximate (1.1)-(1.7) by the following problem :

$$
(3.3) \qquad u_h \in W^{1,\infty}([0,T]; V_h)
$$

$$
(3.4) \quad \ddot{u}_h \in M^1([0, T]; V_h), \text{ the set of measures on } [0, T] \text{ with values in } V_h
$$

$$
(3.5) \qquad u_h(0) = u_h 0
$$

$$
(u_{h1} - \lim_{h \to 0} \dot{u}_h(h), v_h - u_h(0)) \le 0, \forall v_h \in K_h
$$

$$
(3.6) \qquad \left| u_{h1} \right| = \left| \dot{u}_h (0 + 0) \right|
$$

$$
(3.7) \qquad \left\{
\begin{array}{l}
(\ddot{u}_h(t), v_h - u_h(t)) + a(u_h(t), v_h - u_h(t)) \ge \\
\ge (f, v_h - u_h(t)), \forall v_h \in K_h \\
u_h(t) \in K_h, \forall t
\end{array}
\right.
$$

where $u_{h0}$ and $u_{h1}$ are selected in such a way that

$$
\lim_{h \to 0} (\| u_{h0} - u_0 \| + \| u_{h1} - u_1 \|) = 0
$$

If we define a linear operator $A_h$ in $V_h$ by

$$
(3.8) \qquad (A_h u_h, v_h) = a(u_h, v_h), \forall v_h \in V_h
$$

and a maximal monotone graph $\partial\psi_{K_h}$ (see [2], [8]) by

$$
(3.9) \quad \partial_{\psi_{K_h}} (u_h) = \begin{cases} 0 & \text{if } u_h \in K_h \\ \{x/(x, v_h - u_h) \le 0, \forall v_h \in K_h\}, & \text{if } u_h \in \partial K_h \\ \emptyset & \text{elsewhere,} \end{cases}
$$
then (3.7) can be written as

$$ \ddot{u}_h + A_h u_h + \partial\psi_{K_h}(u_h) \ni f_h, $$

where $f_h(t)$ is the projection of $f(t)$ on $V_h$ for the scalar product of $H$. We know from [9] that problem (3.3)-(3.7) possesses a (not necessarily unique) solution which satisfies the energy relation

$$ \begin{align} |\dot{u}_h(t \pm 0)| + a(u_h(t), u_h(t)) &= \tag{3.10} \\ &= |u_{h1}|^2 + a(u_{h0}, u_{h0}) + 2 \int_0^t (f(s), \dot{u}_h(s)) ds \nonumber \end{align} $$

Observe that the left hand side of (3.10) makes sense because $u_h$ is of bounded variation. We call $u_h$ an arbitrarily chosen solution of (3.3)-(3.7) which satisfies (3.10).

**Theorem 3.1** - The sequence $u_h$ converges in the strong topology of $X_2$ to the unique solution of (2.26)-(2.28).

**Proof** - From relation (3.10), a standard Gronwall inequality implies that

$$ (3.11) \qquad \sup_h \esssup_{[0,T]} (|\dot{u}_h(t)| + \|u_h(t)\|) < +\infty $$

Therefore, we may extract a subsequence, still denoted by $u_h$ such that

$$ (3.12) \qquad \left\{ \begin{array}{ll} u_h \to u \quad \text{in } L^\infty(0,T; V) & \text{weak} \\ u_h \to u \quad \text{in } L^\infty(0,T; H) & \text{weak} \end{array} \right. $$

and, thanks to Lemma 2.1

$$ (3.13) \qquad u_h \to u \quad \text{in } C^{0,\alpha}(Q_T), \forall \alpha < \frac{1}{2} $$
In particular, we can see that

$$ (3.18) \qquad u \in \mathcal{K}. $$

We cannot use arbitrary test functions in $\mathcal{K}$, and we will need an approximation.

Let $v$ be an element of $\mathcal{K}$ which is equal to $u$ for $t \ge T - \varepsilon$; for $\eta \le \varepsilon/2$, define

$$ (3.19) \qquad v^{\eta}(x, t) = \begin{cases} u(x, t) + \eta^{-1} \int_{t}^{t+\eta} (v-u)(x,t) dx + \\ + k(\eta)x\psi(t) & \text{if } t \le T - \eta \\ u(x,t) & \text{if } t \ge T - \eta \end{cases} $$

where $\psi$ is a non negative smooth function of $t$, equal to $1$ on $[0, T - \varepsilon]$ and to zero on $[T - \varepsilon/2, T]$, and $k(\eta)$ will be chosen later.

As $u$ belongs to $C^0, \frac{1}{2} (\bar{Q}_T)$,

$$ \begin{aligned} & |u(L,t) - \eta^{-1} \int_t^{T+\eta} u(L,s) ds| \le \eta^{-1} \int_t^{T+\eta} |u(L,t) - u(L,s)| ds \le \\ & \le C \|u\|_2 \eta^{-1} \int_0^\eta \sqrt{s} \, ds = 2C \|u\|_2 \sqrt{\eta}/3. \end{aligned} $$

We choose

$$ k(\eta) = 2C \|u\|_2 \sqrt{\eta} \cdot L^{-1}, $$

and we will be sure that

$$ (3.20) \qquad v^{\eta}(L,t) \ge k_0 + C \|u\|_2 \sqrt{\eta}, \text{ for } t \le T - \varepsilon. $$

On $[T - \varepsilon, T]$,

$$ v^{\eta}(x,t) = u(x,t) + k(\eta)x\psi(t) $$
and $v^η(t)$ still belongs to $K$. We know from (3.12) that $u$ belongs to $X_∞$; on the other hand,

$$ \| (v^η - u)(t) \| \le k(η) \psi(t) L + η^{-\frac{1}{2}} \| v - u \|_2, $$

so that

$$ (3.21) \qquad v^η \in L^\infty(0, T; V). $$

If we denote by $Q_h$ the projection onto $V_h$ which is orthogonal with respect to the scalar product $((\cdot, \cdot))$ of $V$, condition (3.1) implies

$Q_h \to I$ in the strong operator topology, and thus, thanks to the Sobolev inclusion theorem,

$$ \| Q_h z - z \|_{C^0} \le C(h) \| z \| $$

$$ \forall z \in V, $$

with

$$ \lim_{h} C(h) = 0. $$

In particular, we have

$$ | (Q_h v^η)(L,t) - v^η(L,t) | \le C(h) \quad \text{ess sup} \quad \| v^η(t) \|_{t \in [0,T]} $$

so that, thanks to (3.13), (3.20) and (3.21),

$$ \begin{gathered} [u_h + Q_h(v^\eta - u)](t) \in K_h, \forall t \in [0, T], \tag{3.22} \\ \text{and for all large enough } n. \end{gathered} $$

The test function we shall use in (3.7) is given by

$$ v_h = u_h(t) + Q_h(v^\eta(t) - u(t)); $$
We integrate the inequality on $[0, T]$; we may integrate by parts the first term, because $\dot{v}^h - \dot{u}$ is in $\mathcal{X}_2$, by construction. We obtain :

$$ (3.23) \quad \left\{ \begin{aligned} & -(\dot{u}_h(0+0), Q_h(v_h(0)-u(0))) - \int_0^T (u_h(t), Q_h[\dot{v}_h(t)-\dot{u}_h(t)]) dt \\ & + \int_0^T a(u_h(t), Q_h(v_h(t)-u_h(t))) dt \ge \int_0^T (f(t), Q_h(v_h(t)-u_h(t))) n \end{aligned} \right. $$

From (3.1) and (3.6),

$$ \lim_{h \to 0} \dot{u}_h(0+0) = u_1 $$

We may easily pass to the limit in all the other terms, using the convergences in the strong topology of $V$:

$$ Q_h(\dot{v}_h(t) - \dot{u}_h(t)) \to \dot{v}^n(t) - \dot{u}_h(t), \forall t $$

$$ Q_h(v_h(t) - u_h(t)) \to v^n(t) - u(t), \forall t $$

We pass to the limit with respect to $\eta$ in the expression we obtained, and we get (2.28).

We have an energy equality for $u$ solution of (2.26) - (2.28), by integrating

$$ (u_t^2 + u_x^2)_t - 2(u_x u_t)_t = 2 f u_x $$

on $[0, L] \times [0, \tau]$:

$$ (3.24) \quad \left\{ \begin{aligned} & \int_0^L [u_t(x, \tau)^2 + u_x(x, \tau)^2] dx = \int_0^L [u_1(x)^2 + (du_0/dx)(x)^2] dx \\ & + 2 \int_0^\tau \int_0^L f u_x dx dt \end{aligned} \right. $$
as the boundary terms disappear thanks to the Dirichlet condition (1.1) on the left and the
unilateral conditions (1.3)-(1.5) on the right. An immediate comparison of (3.10) and
(3.23) shows that, with (3.12),

$$
\lim_{h \to 0} \int_0^T (|u_h(t)|^2 + ||u_h(t)||^2) dt = \int_0^T (|u(t)|^2 + ||u(t)||^2) dt.
$$

As $\mathcal{X}_2$ is endowed with a Hilbert structure, as $u_h$ converges weakly to $u$ in $\mathcal{X}_2$, and $||u_h||_2$
converges to $||u||_2$, we conclude that $u_n$ converges strongly to $u$ in $\mathcal{X}_2$, and this ends the
proof of theorem 3.1. ■

III.2 - CONVERGENCE OF FULLY DISCRETIZED FINITE ELEMENTS SCHEMES,

Given $V$ as in (3.1), we define as fully discretized scheme

$$
(3.25) \quad \left\{
\begin{array}{ll}
\displaystyle \left( \frac{u_h^{k+1} - 2u_h^k + u_h^{k-1}}{\Delta t^2} , v - u_h^{k+1} \right) + & \\
\displaystyle \quad + a(u_h^k, v - u_h^{k+1}) \ge (f_h^k, v - u_h^{k+1}), \quad \forall v \in K_h, u_h^{k+1} \in K_h & \\
\text{with initial conditions } u_h^0 \text{ and } u_h^1 \text{ such that} &
\end{array}
\right.
$$

with initial conditions $u_h^0$ and $u_h^1$ such that

$$
u_h^0 \to u_0 \text{ in } V
$$

$$
(3.26) \qquad \Delta t^{-1} (u_h^1 - u_h^0) \to u_t \quad \text{in } H.
$$

and $f_h^k$ a suitable discretization in time of $f_h$.

Relation (3.25) can be written equivalently, using notation (3.9), as

$$
\frac{u_h^{k+1} - 2u_h^k + u_h^{k-1}}{\Delta t^2} + A_h u_h^k + \partial \psi_{K_h} (u_h^{k+1}) \ni f_h^k,
$$
We shall say therefore that it is implicit in the constraint, and it is quite clear that (3.25) defines exactly one $u^k$ at each time step.

**Theorem 3.2** The scheme (3.25) converges as $h \to 0$ and $\Delta t \to 0$ under the stability condition

$$ (3.27) \qquad \Delta t \le 2 \left[ \frac{(1-\alpha)}{\kappa_h} \right]^{\frac{1}{2}} $$

where $\alpha$ is an arbitrarily small positive number and

$$ \kappa_h = \sup_{\substack{u \in V \\ u \neq 0}} \left[ \frac{(A_h u, u)}{|u|^2} \right]^{\frac{1}{2}} $$

**Proof** - Let us first prove the stability : if we take $v = u_h^{k-1}$ in (3.25), we can write (3.25) as

$$ (3.28) \qquad \begin{aligned} & a(u_h^k, u_h^{k+1}) + \left| \frac{u_h^{k+1} - u_h^k}{\Delta t} \right|^2 \le a(u_h^k, u_h^{k-1}) + \\ & \qquad + \left| \frac{u_h^k - u_h^{k-1}}{\Delta t} \right|^2 + (f_h^k, u_h^{k+1} - u_h^{k-1}). \end{aligned} $$

Let us show that

$$ (3.29) \qquad R(u, v) = (A_h u, v) + \left| \frac{u - v}{\Delta t} \right|^2 \ge \frac{1}{4} \|u + v\|^2 + \alpha \left| \frac{u - v}{\Delta t} \right|^2 $$

We have the identity

$$ (A_h u, v) = \frac{1}{4} (A_h (u+v), u+v) - \frac{1}{4} (A_h (u-v), u-v) $$

and, thus,

$$ \begin{aligned} R(u, v) &\ge \frac{1}{4} (A_h (u+v), u+v) + \left| \frac{u-v}{\Delta t} \right|^2 \left( 1 - \frac{\| \kappa_h \| \Delta t^2}{4} \right) \\ &\ge \frac{1}{4} (A_h (u+v), (u+v)) + \alpha \left| \frac{u-v}{\Delta t} \right|^2, \end{aligned} $$

according to (3.27).
If we perform a discrete integration on (3.28), we obtain

$$ 
\begin{aligned}
& a(u_h^k, u_h^{k+1}) + \left| \frac{u_h^{k+1} - u_h^k}{\Delta t} \right|^2 \\
\le & a(u_h^0, u_h^1) + \left| \frac{u_h^1 - u_h^0}{\Delta t} \right|^2 + \sum_{\ell=1}^{k} (f_h^\ell, u_h^{\ell+1} - u_h^{\ell-1}),
\end{aligned}
$$

which we estimate by

$$ (3.30) \qquad \alpha \left| \frac{u_h^{k+1} - u_h^k}{\Delta t} \right|^2 \le C_h + \sum_{\ell=0}^{k} \left| \frac{u_h^{\ell+1} - u_h^\ell}{\Delta t} \right|^2 \Delta t, $$

where

$$ C_h = a(u_h^0, u_h^1) + \left| \frac{u_h^1 - u_h^0}{\Delta t} \right|^2 + \sum_{\ell=1}^{M} |f_h^\ell|^2 \Delta t $$

and $M = M(\Delta t)$ is defined by

$$ T/\Delta t \le M < 1 + (T/\Delta t) $$

Now we have a Lemma on discrete Gronwall inequalities :

**Lemma 3.2** - Let $x_k$ be a sequence of numbers which satisfies

$$ (3.31) \qquad x_k \le a + b \sum_{\ell=0}^{k} x_\ell \Delta t. $$

Then, we have the following estimate on $x_\ell$ :

$$ (3.32) \qquad x_k \le (a + b x_0 \Delta t)(1 - b \Delta t)^{-k} $$

*Proof* - Classical; see for instance [11], Lemma 4.1, page 76.
We apply Lemma 3.2 to (3.30) with $x_k = \left| \frac{u_h^{k+1} - u_h^k}{\Delta t} \right|^2$, $a = C_h \alpha^{-1}$, $b = \alpha^{-1}$, and obtain, using (3.29)

$$ \frac{1}{4} \| u_h^k + u_h^{k+1} \|^2 + \alpha \left| \frac{u_h^{k+1} - u_h^k}{\Delta t} \right|^2 \le (C_h + \left| \frac{u_h^1 - u_h^0}{\Delta t} \right|^2 \Delta t) \alpha^{-1} (1 - \alpha^{-1} \Delta t)^{-k} $$

As $C_h$ is bounded independently of $h$, and as we have the estimate, for $v$ in $V_h$

$$ (3.33) \qquad |v/\Delta t|^2 \ge [4(1-\alpha)]^{-1} \|v\|^2, $$

which is a consequence of (3.27), we get

$$ (3.34) \qquad \left\{ \begin{array}{l} \displaystyle \max_{h,k} \left| \frac{u_h^{k+1} - u_h^k}{\Delta t} \right|, \|u_h^k\| \text{ is bounded independently of } h, \\[1em] \text{and of } k \le M. \end{array} \right. $$

Define (with the time step understated) a function $\|\cdot\|_h$ by

$$ \begin{gather*} u_h(x, t) = u_h^k(x) \frac{(k+1)\Delta t - t}{\Delta t} + u_h^{k+1}(x) \frac{t - k \Delta t}{\Delta t} \\ \text{for } k \Delta t \le t \le (k+1) \Delta t. \end{gather*} $$

Relation (3.34) implies that we can extract a subsequence of the sequence of the $U$'s such that

$$ u_h \to u \text{ in } L^{\infty}(0, T; V) \quad \text{weak } * $$

$$ \frac{dU_h}{dt} \to u \text{ in } L^{\infty}(0, T; H) \quad \text{weak } * $$

$$ u_h \to u \text{ in } C^0(\bar{Q}_T), \forall \alpha < \frac{1}{2} $$
If we choose $v^7$ as in (3.19), and let

$$v_h^k = u_h^{k+1} + Q_h(v_h^7(k\Delta t) - u_h(k\Delta t)),$$

if we replace $v$ by the defined $v_h^k$ in (3.25), and perform a discrete integration in time, we obtain

$$ 
\begin{aligned}
& - \left( \frac{u_h^1 - u_h^0}{\Delta t} , V_h^0 - U_h^1 \right) - \\
& \qquad M - 1 \sum_{k=1}^{M-1} \left( \frac{u_h^k - u_h^{k-1}}{\Delta t} , \frac{v_h^k - u_h^{k+1} - v_h^{k-1} + u_h^k}{\Delta t} \right) \Delta t + \\
& \qquad + \sum_{k=1}^{M-1} a_h^k (u_h^k, v_h^k - u_h^{k+1}) \Delta t \ge \\
& \qquad \ge \sum_{k=1}^{M-1} (f_h^k, v_h^k - u_h^{k+1}) \Delta t.
\end{aligned}
$$

We pass to the limit for $\Delta t \to 0$ and $h \to 0$ in this expression, to obtain (2.28). To see that $U_h$ converges strongly in $X_2$ to $u$, it is enough to show that the total energy of $U_h$ converges to the total energy of $u$. Relation (3.28) implies that

$$ \lim_{n \to \infty} \int_{Q_T} \left( \left| \frac{\partial U_h}{\partial x} \right|^2 + \left| \frac{\partial U_h}{\partial t} \right|^2 \right) dx dt < \int_{Q_T} \left( \left| \frac{\partial u}{\partial x} \right|^2 + \left| \frac{\partial u}{\partial t} \right|^2 \right) dx dt $$

The weak convergence of $U$ to $u$ has as a consequence that

$$ \lim_{n \to \infty} \int_{Q_T} \left( \left| \frac{\partial U_h}{\partial x} \right|^2 + \left| \frac{\partial U_h}{\partial t} \right|^2 \right) dx dt \ge \int_{Q_T} \left( \left| \frac{\partial u}{\partial x} \right|^2 + \left| \frac{\partial u}{\partial t} \right|^2 \right) dx dt. $$

and this completes the proof of the theorem. ■
Let us now compute the value of $K_h$ for P1 uniform finite elements, corresponding to a step $\Delta x = L/n$.

If we take P1 finite elements, and if $u$ is affine on intervals $[j\Delta x, (j+1)\Delta x]$, then

$$ (A_h u, u) = \sum_{j=0}^{n-1} \left| \frac{u((j+1)\Delta x) - u(j\Delta x)}{\Delta x} \right|^2 \Delta x $$

and

$$ |u|^2 = \Delta t \left| \sum_{j=1}^{n-1} 2u(j\Delta x)^2 + u(n\Delta x)^2 + \sum_{j=1}^{n-1} u(j\Delta x)u(j+1)\Delta x \right| $$

so that

$$ (A_h u, u) \le 12 \Delta x^{-2} |u|^2 $$

and $K_h = 12/\Delta x^2$, as the reverse inequality can be checked easily.

# IV - EXPLICIT DIFFERENCE SCHEMES.

## IV. 1 - NOTATIONS AND DESCRIPTION OF THE FAMILY OF SCHEMES.

Let $n$ be an integer, let $h = \Delta x = L/n$ be the space step, and $\Delta t$ be the time. We denote by $u_j^k$ the finite difference scheme defined as follows:

$$ (4.1) \qquad \left\{ \begin{array}{ll} \displaystyle \eta_j^{k+1} = 2\eta_j^k - \eta_j^{k-1} + \frac{\Delta t^2}{\Delta x^2} (\eta_{j+1}^k - 2\eta_j^k + \eta_{j-1}^k) + \Delta t^2 F_j^k & \text{for } 1 \le j \le n-1 \\ \\ \eta_n^{k+1} = \max [2\eta_n^k - \eta_n^{k-1} + \frac{\Delta t^2}{\Delta x^2} (\eta_{n-1}^k - \eta_n^k) + \Delta t^2 F_j^k, k=0] & \text{for } n \le j \le n+1 \end{array} \right. $$

where $F_j^k$ is a suitable discretization of $f$. Whenever necessary $\eta_0^k = 0$
If we let

$$
(4.2) \qquad u_h^k(x) = \sum_{j=1}^{n} u_j^k g\left(\frac{x - j \Delta x}{\Delta x}\right)
$$

where

$$
(4.3) \qquad g(x) = \max (1 - |x|, 0),
$$

then $u_h^k$ belongs to $V_h$, the space of piecewise affine function on intervals $[jh, (j + 1)h]$.

Moreover, for $u$ and $v$ in $V_h$, with $u = \sum_{j=1}^{n} u_j f(\frac{x-j\Delta x}{\Delta x})$, $v = \sum_{j=1}^{n} v_j f(\frac{x-j\Delta x}{\Delta x})$,
let us define an alternative scalar product on $V_h$ by

$$
(4.4) \qquad s_h(u, v) \coloneqq \sum_{j=1}^{n} u_j v_j \Delta x .
$$

We can give now an equivalent variational form of (4.1).

$$
u_h^{k+1} \in K_h = K \cap V_h
$$

$$
(4.5) \qquad s_h \left( \frac{u_h^{k+1} - 2u_h^k + u_h^{k-1}}{\Delta t^2} , v - u_h^{k+1} \right) + a(u_h^k, v - u_h^{k+1}) \ge
$$
$$
\ge s_h(f_h^{k+1}, v - u_h^{k+1}) , \forall v \in K_h
$$

The equivalence of (4.1) and (4.5) can be readily checked, as follows ;

$$
K_h = |u \in V_h / u_n \ge k_0 |
$$

If in (4.5), $V_h = \mathcal{V}_h^{k+1} + \Phi_h$, then

$$
s_h \left( \frac{u_h^{k+1} - 2u_h^k + u_h^{k-1}}{\Delta t^2}, \varphi \right) + a(u_h^k, \varphi) \ge s_h(f_h^{k+1}, \varphi), \forall \varphi \in V_h \text{ such that } \Phi_n = 0,
$$
which is clearly equivalent to the first part of (4.1). If $V_j = 0, \forall j \le n-1$, in (4.5), we get

$$ 
\begin{aligned}
& \left( \frac{u_n^{k+1} - 2u_n^k + u_n^{k-1}}{\Delta t^2} \right) (v_n - u_n^{k+1}) + \frac{u_n^k - u_{n-1}^k}{\Delta x^2} (v_n - u_h^{k+1}) \ge \\
& \ge \Delta t^2 F_n^k (v_n - u_n^{k+1}), \forall v_n \ge k_0,
\end{aligned}
$$

and it is straight forward to check that this is precisely equivalent to the second part of (4.1).

To make things more precise, we shall settle for the following approximations of the exterior forces, and of the initial data :

$$ (4.6) \qquad \dot{u}_j^0 = u_0 (j \Delta x) $$

$$ (4.7) \qquad \left\{ \begin{array}{l} \displaystyle \dot{u}_j^1 = \dot{u}_j^0 + \frac{\Delta t}{\Delta x} \int_{(j-\frac{1}{2})\Delta x}^{(j+\frac{1}{2})\Delta x} u_1(s) ds, \text{ if } 1 \le j \le n-1 \\ \\ \displaystyle \dot{u}_n^1 = \dot{u}_n^0 + \frac{2\Delta t}{\Delta x} \int_{(n-\frac{1}{2})\Delta x}^{n\Delta x} u_1(s) ds \end{array} \right. $$

$$ (4.8) \qquad F_j^k = \frac{1}{\Delta t \Delta x} \int_{(j-1)\Delta x}^{j\Delta x} \int_{k\Delta t}^{(k+1)\Delta t} f(x, t) \, dx \, dt $$

We shall consider in the remainder of this chapter that $\lambda = \Delta t / \Delta x$ is fixed in $[0, 1]$, and we shall prove the convergence of the scheme, first in the non-characteristic case $\lambda < 1$.

## IV. II - CONVERGENCE IN THE NON CHARACTERISTIC CASE.

**Proposition 4.1** - Let $\lambda$ be less than 1 : Then, we have the following estimate on $u_h^k$ :

$$ (4.9) \qquad \| u_h^k \| \text{ and } s_h \left( \frac{u_h^{k+1} - u_h^k}{\Delta t}, \frac{u_h^{k+1} - u_h^k}{\Delta t} \right)^{\frac{1}{2}} \text{ are bounded independently of } k \text{ and } \Delta x. $$
Proof - Take $v = u_h^{k-1}$ in (4.5); then, we have

$$
\begin{align*}
& a(u_h^k, u_h^{k+1}) + s_h \left( \frac{u_h^{k+1} - u_h^k}{\Delta t}, \frac{u_h^{k+1} - u_h^k}{\Delta t} \right) \le \\
& \le a(u_h^k, u_h^{k-1}) + s_h \left( \frac{u_h^k - u_h^{k+1}}{\Delta t}, \frac{u_h^k - u_h^{k-1}}{\Delta t} \right) + \\
& \quad + s_h \left( f_h^k, \frac{u_h^{k+1} - u_h^{k-1}}{\Delta t} \right).
\end{align*}
$$

If we add these inequality for $k = 1$ to $l$, we have

$$
(4.10) \quad \left\{
\begin{aligned}
& a(u_h^l, u_h^{l+1}) + s_h \left( \frac{u_h^{l+1} - u_h^l}{\Delta t}, \frac{u_h^{l+1} - u_h^l}{\Delta t} \right) \le \\
& \le a(u_h^1, u_h^0) + s_h \left( \frac{u_h^1 - u_h^0}{\Delta t}, \frac{u_h^1 - u_h^0}{\Delta t} \right) + \\
& \qquad \sum_{k=1}^{l} s_h \left( f_h^k, \frac{u_h^{k+1} - u_h^{k-1}}{\Delta t} \right) \Delta t
\end{aligned}
\right.
$$

We estimate $a$ in terms of $s_h$ on $V_h$:

$$
a(u, u) = \sum_{j=0}^{n-1} \left( \frac{u_{j+1} - u_j}{\Delta x} \right)^2 \Delta x \le
$$

$$
\begin{align*}
& \le \sum_{j=0}^{n-1} \frac{2(u_{j+1}^2 + u_j^2)}{\Delta x} \le 4 \sum_{j=0}^{n-1} u_j^2 \Delta x \Delta x^{-2}, \\
& \text{so that}
\end{align*}
$$

so that

$$
(4.11) \qquad a(u, u) \le 4 \Delta x^{-2l} s_h(u, u), \forall u \in V_h.
$$
Therefore,

$$
\begin{align*}
& a(u, v) + s_h \left( \frac{u - v}{\Delta t}, \frac{u - v}{\Delta t} \right) = \\
&= \frac{1}{4} a(u + v, u + v) - \frac{1}{4} a(u - v, u - v) + s_h \left( \frac{u - v}{\Delta t}, \frac{u - v}{\Delta t} \right) \\
&\ge \frac{1}{4} a(u + v, u + v) + s_h \left( \frac{u - v}{\Delta t}, \frac{u - v}{\Delta t} \right) (1 - \lambda^2),
\end{align*}
$$

and, in particular

$$
(4.12) \quad a(u,v) + s_h \left( \frac{u-v}{\Delta t}, \frac{u-v}{\Delta t} \right) > (1-\lambda^2) s_h \left( \frac{u-v}{\Delta t}, \frac{u-v}{\Delta t} \right)
$$

If we use (4.12) in (4.10), we obtain

$$
(1 - \lambda^2) s_h \left( \frac{u_{h}^{l+1} - u_{h}^{l}}{\Delta t}, \frac{u_{h}^{l+1} - u_{h}^{l}}{\Delta t} \right) \le
$$
$$
\le C_h + \sum_{k=0}^{l} s_h \left( \frac{u_{h}^{k+1} - u_{h}^{k}}{\Delta t}, \frac{u_{h}^{k+1} - u_{h}^{k}}{\Delta t} \right)^2 \Delta t,
$$

where

$$
C_h = a(u^1, u^0) + s_h \left( \frac{u_h^1 - u_h^0}{\Delta t}, \frac{u_h^1 - u_h^0}{\Delta t} \right) + \\
M \\
+ \sum_{k=1}^{\infty} s_h (f^k, f^k) \Delta t
$$

and

$$
T/\Delta t \le M \le (T/\Delta t) + 1 .
$$

Then, we apply the discrete Gronwall Lemma stated as Lemma 3.2; we take in (3.31)
$a = (1 - \lambda^2)^{-1} C_h, b = (1 - \lambda^2)^{-1},$ and obtain
$$
\begin{equation} \tag{4.13}
\begin{aligned}
& \frac{1}{4} a \left(u_h^{k+1} + u_h^k u_{h,0}^{k+1} + u_h^k u_{h,0}^k\right) + (1-\lambda^2) s_h \left(\frac{u_h^{k+1} - u_h^k}{\Delta t}, \frac{u_h^{k+1} - u_h^k}{\Delta t}\right) \ll \\
& \qquad \le \left(C_h + \left|\frac{u_{h,0}^1 - u_h^0}{\Delta t}\right|^2 \Delta t \right) (1-\lambda^2)^{-1} (1 - (1-\lambda^2)\Delta t)^{-k}
\end{aligned}
\end{equation}
$$

Then, with the help of (4.11), and the fact that $C_h$ is bounded independently of $h$, we obtain
(4.9). ■

We interpolate $u_h^k$ by letting

$$
(4.14) \quad u_h(x,t) = u_h^k(x) \frac{(k+1)\Delta t - t}{\Delta t} + u_h^{k+1}(x) \frac{t - k\Delta t}{\Delta t}.
$$

Then, we have the following convergence result :

Theorem 4.2 - The sequence $u_h$ converges to the solution of (2.26) - (2.28) in the strong topology of $X_2$.

Proof - The difference with the corresponding finite elements proof (theorem 3.2) is that we
have a scalar product $s_h$, defined on $V_h$ by (4.4). From the obvious estimate

$$
|u| \le s_h (u, u)^{\frac{1}{2}}
$$

and Proposition 4.1, we deduce that we can extract a converging subsequence still denoted
by $u_h$ such that

$$
u_h \to u \text{ in } L^\infty(0, T; H) \text{ weak*}
$$

$$
\frac{\partial u_h}{\partial t} \to \frac{\partial u}{\partial t} \quad \text{in } L^\infty(0, T; V) \text{ weak*}
$$

$$
u_h \to u \text{ in } C^0(\bar{Q}).
$$
Clearly, $u$ belongs to $K$. For any $v$ in $K$, such that $v = u$ for large enough time, we define $v^n$ as in (3.19), and we let

$$v_h^k = u_h^{k+1} + Q_h (v_h^n (k\Delta t) - u_h (k\Delta t)).$$

If we substitute $v$ by $v_h^k$ in (4.5), and perform a discrete integration, with an integration by parts on the first term we get

$$ (4.15) \quad \left\{ \begin{array}{@{}l@{\quad}l@{}} & -s_h \left( \frac{u_h^1 - u_h^0}{\Delta t}, v^0 - u^1 \right) - \sum_{k=1}^{M-1} s_h \left( \frac{u_h^k - u_h^{k-1}}{\Delta t}, \frac{v_h^k - u_h^{k+1} - v_h^{k-1}}{\Delta t} + u_h^k \right) \Delta t \\ & + \sum_{k=1}^{M-1} a_h (u_h^k, v_h^k - u_h^{k+1}) \Delta t \ge \sum_{k=1}^{M-1} s_h (v_h^k, u_h^k - u_h^{k+1}) \Delta t. \end{array} \right. $$

Let $\psi$ belong to $V$, and let

$$\psi_h = Q_h \psi$$

$$\psi_h = \sum_{j=1}^{n} \psi_j \left(1 - \left| \frac{x - j\Delta x}{\Delta x} \right|\right)^+.$$

Then, for any $v_h$ in $V_h$,

$$ \begin{align*} s_h(v_h, \psi_h) - (v_h, \psi_h) &= \sum_{j=1}^{n} v_j \psi_j \Delta x - \frac{2}{3} \sum_{j=1}^{n} v_j \psi_j \Delta x - \frac{1}{3} \sum_{j=1}^{n} v_j \psi_j \Delta x - \\ & \quad - \frac{1}{6} \sum_{j=1}^{n} \left( v_j \psi_{j+1} + v_{j+1} \psi_j \right) \Delta x = \\ & \quad = \sum_{j=1}^{n} v_j \left( \frac{1}{3} \psi_j - \frac{1}{6} \psi_{j-1} - \frac{1}{6} \psi_{j+1} \right) \Delta x + \\ & \quad + v_n \left( \frac{2}{3} \psi_n - \frac{1}{6} \psi_{n-1} \right) \Delta x \end{align*} $$
Now, we can see that

$$ (4.16) \quad |s_h(v_h, \phi_h) - (v_h, \phi_h)| \le C (|v_h| |\psi_h| \Delta x + |v_h| |\psi_h| \sqrt{\Delta x}) . $$

where C does not depend on h.

If we replace in (4.15), all the terms involving $s_h$ by similar terms involving the original scalar product in $\mathcal{H}$, the error we commit is estimated by

$$ C \sqrt{\Delta x} \left[ \begin{aligned} & \frac{u_h^1 - u_h^0}{\Delta t} \| Q_h (v_h^\eta (0) - u(0)) \| + \\ & \qquad + \sum_{k=1}^{M-1} \Delta t \left| \frac{u_h^k - u_h^{k-1}}{\Delta t} \| Q_h (v_h^\eta (k\Delta t) - u(k\Delta t) - v_h^\eta ((k-1)\Delta t)) \| + \\ & \qquad + u(k-1)\Delta t \| + \sum_{k=1}^{M-1} \Delta t \left| f_h^k \| Q_h (v_h^\eta (k\Delta t) - u(k\Delta t)) \| \end{aligned} \right] $$

which tends to zero as $h$ tends to zero. We are therefore reduced to the proof of the finite element case, and conclusion is clear. ■

### IV. 3 - CONVERGENCE IN THE CHARACTERISTIC CASE.

The characteristic case is technically more difficult, because the energy estimate involves only the characteristic derivatives, and the fact that we have two superposed grids, which are coupled only at the boundary $x = L$ is important in the definition of a good interpolation.

**Proposition 4.2** - Let $\lambda = 1$, and define the energy at time $k$ by

$$ \mathcal{E}_h^k = (\Delta x)^{-1} \sum_{i=0}^{n-1} |u_{i+1}^k - u_i^{k+1}|^2 + \sum_{j=0}^{n-1} |u_j^k - u_{j+1}^{k-1}|^2 + |u_h^k - u_n^{k-1}|^2 $$

Then,

$$ (4.17) \qquad \mathcal{E}_h^k \text{ is bounded independently of } h \text{ and } k. $$
Proof - Relation (4.1) implies that

$$ U_{j+1}^{k+1} - U_j^k = U_{j+2}^k - U_{j+1}^{k-1} + F_{j+1}^k \Delta x^2 \quad \text{if } 0 \le j \le n-2 $$

$$ U_{j+1}^{k+1} - U_{j+1}^{k} = U_{j-1}^{k} - U_{j-1}^{k-1} + F_{j}^{k} \Delta x^{2} \quad \text{if } 1 \le j \le n-1 $$

Therefore,

$$ (4.18) \quad \left\{ \begin{array}{ll} \sum_{j=0}^{n-2} |U_{j+1}^{k+1} - U_j^k|^2 = \sum_{j=1}^{n-1} |U_{j+1}^k - U_j^{k-1}|^2 + F_j^k \Delta x^2 & \text{and} \\ \\ \sum_{j=1}^{n-1} |U_j^{k+1} - U_{j+1}^k|^2 = \sum_{j=0}^{n-1} |U_j^k - U_{j+1}^{k-1}|^2 + F_{j+1}^k \Delta x^2 & \text{and} \end{array} \right. $$

The remaining terms are

$$ \begin{aligned} & |U_n^{k+1} - U_{n-1}^k|^2 + |U_0^{k+1} - U_1^k|^2 + |U_n^{k+1} - U_n^k|^2 - \\ & \quad - |U_1^k - U_{n-1}^{k-1}|^2 - |U_n^k - U_{n-1}^{k-1}|^2 - |U_n^k - U_n^{k-1}|^2. \end{aligned} $$

Observe first that $U_0^{k-1} = U_0^k = U_{n-1}^k = 0$, so that

$$ (4.19) \qquad |U_0^{k+1} - U_1^k|^2 = |U_1^k - U_0^{k-1}|^2. $$

Let now

$$ (4.20) \qquad U_n^k = a \qquad U_{n-1}^{k-1} = b \qquad U_n^{k+1} = c \qquad U_{n-1}^k = d $$

We know that

$$ a \ge k_0, b \ge k_0, c \ge k_0, $$
and if

$$g = \Delta x^2 F_n^k,$$

Then, according to (4.1),

$$c = \max (a - b + d + g, k_0).$$

We must estimate $|c - d|^2 + |c - a|^2$ in terms of $a, b, d, g$. We have the two following identities

$$|e - d|^2 + |c - a|^2 - |b - d|^2 - |b - a|^2 = 2(c - b)(c + b - d - a)$$

$$|c - d|^2 + |c - a|^2 - |b - g - d|^2 - |b - g - a|^2 = 2(c - b + g)(c + b - d - a - g)$$

Therefore, if $a - b + d + g \ge k$,

$$c = a - b + d + g,$$

so that

$$|c - d|^2 + |c - a|^2 \le |b - g - d|^2 + |b - g - a|^2.$$

Assume now that $a - b + d + g \le k$. Then

$$c = k, c \le b \text{ and } a - b + d + g \le c,$$

so that

$$c + b - d - a - g \ge 0.$$

Therefore, if

$$c \le b - g, \text{ (i.e. } k \le b - g),$$

we have

$$(c - b + g)(c + b - d - a - g) \le 0.$$
If $k \ge b - g$, then, as $b \ge k$, we have $g \ge 0$, so that

$$c + b - a - d \ge 0 \text{ and } c < b.$$

It follows that

$$(c - b)(c + b - d - a) \le 0,$$

and, finally, in all cases

$$(4.21) \quad |c - d|^2 + |c - a|^2 \le \max(|b - d|^2 + |b - a|^2, |b - g - d|^2 + |b - g - a|^2)$$

We deduce from (4.18) - (4.21) that

$$ \begin{align*} \mathcal{E}^{n+1} &\le \sum_{j=1}^{n-1} |\mathfrak{U}_{j+1}^k - \mathfrak{U}_i^{k-1}| + F_j^k \Delta x^2 | + \\ &\quad + \sum_{j=0}^{n-2} |\mathfrak{U}_j^k - \mathfrak{U}_{j+1}^{k-1}| + F_{j+1}^k \Delta x^2 | + |\mathfrak{U}_1^k - \mathfrak{U}_0^{k-1}|^2 + \\ &\quad + \max_j | |\mathfrak{U}_n^{k-1} - \mathfrak{U}_{n-1}^k|^2 + |\mathfrak{U}_n^{k-1} - \mathfrak{U}_n^k|^2 , |\mathfrak{U}_n^{k-1} - \mathfrak{U}_{n-1}^k| + F_n^k \Delta x^2 |^2 + \\ &\quad + |\mathfrak{U}_n^{k-1} - \mathfrak{U}_n^k| + F_n^k \Delta x^2 |^2 \} \le \\ &\le \mathcal{E}^n \left(1 + \frac{\Delta x}{2}\right) + \sum_{j=1}^n |F_n^k|^2 \Delta x^2 . \end{align*} $$

This is a discrete Gronwall inequality which allows us to conclude. ■

Let us define now a good interpolation of $\mathfrak{U}_j^k$. For this purpose, we define two sets of grid points

$$ (4.22) \quad \left\{ \begin{array}{ll} \bar{\mathcal{G}}_h^k = \{ (jh, kh) / j + k \text{ is even or } j = n \} & \\ \\ \tilde{\mathcal{G}}_h^k = \{ (jh, kh) / j + k \text{ is old or } j = n \} & \end{array} \right. $$
and subspaces of V

$$
(4.23) \quad \left\{
\begin{array}{l}
\bar{v}_h^k = \{ u \in V / u \text{ is affine on } [0, L] \setminus \bar{G}_h^k \} \\
\\
\tilde{v}_h^k = \{ u \in V / u \text{ is affine on } [0, L] \setminus \tilde{G}_h^k \}
\end{array}
\right.
$$

Then we define two interpolations of $\tilde{u}_j^k$, each one corresponding to one subgrid

$$
(4.24) \quad \left\{
\begin{array}{l}
\bar{u}_h^k \in \bar{V}_h^k, \quad u_h^k (j\Delta x) = \bar{u}_j^k, \quad \forall j \in \bar{G}_h^k, \\
\\
\tilde{u}_h^k \in \tilde{V}_h^k, \quad u_h^k (j\Delta x) = \tilde{u}_j^k, \quad \forall j \in \tilde{G}_h^k,
\end{array}
\right.
$$

and

$$
(4.25) \quad \left\{
\begin{aligned}
\bar{u}_h(x,t) &= \bar{u}_h^k(x) \frac{(k+1)\Delta t - t}{\Delta t} + \bar{u}_h^k(x) \frac{t - k\Delta t}{\Delta t} \\
\tilde{u}_h(x,t) &= \tilde{u}_h^k(x) \frac{(k+1)\Delta t - t}{\Delta t} + \tilde{u}_h^k(x) \frac{t - k\Delta t}{\Delta t}
\end{aligned}
\right.
$$

Finally, we let

$$
(4.26) \qquad u_h^{\#}(x,t) = \frac{1}{2} (u_h(x,t) + u_h(x,t)) .
$$

Let us see why $\bar{u}_h$ and $\tilde{u}_h$ are of bounded energy :

$$
(4.27) \quad \frac{\partial u_h^k}{\partial x} = \bar{\Sigma} \frac{u_{j'}^k - u_j^k}{(j' - j) \Delta x} \quad 1_{[j \Delta x, j' \Delta x]}
$$

where the sum is extended to those $j$ and $j'$ in $\bar{G}_h^k$ such that $j' > j$ is the closest neighbor

to $j$. Thus $\frac{\partial \bar{u}_h}{\partial x}$ contains terms of the form $\frac{u_{j+2}^k - u_j^k}{2\Delta x} \mathbf{1}_{[j\Delta x, (j+2)\Delta x]}$ and possibly
terms of the form
$$ \frac{u^k}{\Delta x} \mathbf{1}_{[0, \Delta x]} $$

or

$$ \frac{u_{n}^{k} - u_{n-1}^{k}}{\Delta x} \mathbf{1}_{\{(n-1)\Delta x, n \Delta x\}} $$

From the inequalities

$$ |u_{j+2}^k - u_j^k|^2 \le 2 \left[ (u_{j+2}^k - u_{j+1}^{k-1})^2 + (u_{j+1}^{k-1} - u_j^k)^2 \right], $$

$$ |u_{n}^{k} - u_{n-1}^{k}|^{2} \le 2 | |u_{n}^{k} - u_{n-1}^{k-1}|^{2} + |u_{n}^{k-1} - u_{n-1}^{k}|^{2} | $$

and the fact that

$$ u_1^k = u_1^k - u_0^{k-1} , $$

we can see now that

$$ (4.28) \qquad \left| \frac{\partial \bar{u}_h}{\partial x} \right| (t) \quad \text{is bounded independently of } h, \text{ and } t \text{ in } [0, T] $$

Similarly, if $(jh, kh)$ belongs to $\bar{G}_h$ with $j$ not equal to 0 or $n-1$, we have the following expression of $\frac{\partial \bar{u}_h}{\partial t}$ on the square $((j+1)h, (k+2)h) \times (kh, (k+1)h)$:

$$ \begin{multline} \frac{\partial \bar{u}^h}{\partial t}(x,t) = \frac{u_{j+1}^{k+1} - u_j^k}{\Delta t} + \frac{u_{j+3}^{k+1} - u_{j+1}^{k+1}}{2 \Delta x} \frac{x - (j+1) \Delta x}{\Delta t} - \\ - \frac{u_{j+2}^k - u_j^k}{2 \Delta x} \frac{x - j \Delta x}{\Delta t}, \tag{4.29} \end{multline} $$
with a similar expression on $(jh, (j+1)h) \times (kh, (k+1)h)$. In the end elements, whenever they are needed, $\frac{\partial u}{\partial t}$ takes a particular form, which is the following:

$$
(4.30) \quad \left\{
\begin{aligned}
h \in G_h^k &\Rightarrow \frac{\partial u^h}{\partial t}(x, t) = \frac{u_0^{k+1} - u_0^k}{2 \Delta x} \frac{x}{\Delta t} - \frac{u_1^k - u_0^k}{\Delta x} \frac{x}{\Delta t} \\
h \in G_h^{k+1} &\Rightarrow \frac{\partial u^h}{\partial t}(x, t) = \frac{u_1^{k+1} - u_0^{k+1}}{\Delta x} \frac{x}{\Delta t} - \frac{u_2^k - u_0^k}{\Delta x} \frac{x}{\Delta t}
\end{aligned}
\right.
$$

$$
(4.31) \quad \left\{
\begin{aligned}
& (n-1) h \in G_h^k \Rightarrow \frac{\partial u^h}{\partial t}(x, t) = \frac{u_{n-2}^{k+1} - u_{n-1}^k}{\Lambda t} + \\
& \qquad + \frac{u_n^{k+1} - u_{n-2}^k}{2 \Delta x} - \frac{x - (n-2) \Delta x}{\Delta t} - \frac{u_{n-1}^k - u_{n-2}^k}{\Lambda x} - \frac{x - (n-1) \Delta x}{\Delta t} \\[1em]
& (n-1) h \in G_h^{k+1} \Rightarrow \frac{\partial u^h}{\partial t}(x, t) = \frac{u_{n-1}^{k+1} - u_{n-2}^k}{\Lambda t} + \\
& \qquad + \frac{u_n^{k+1} - u_{n-1}^k}{\Delta x} - \frac{x - (n-1) \Delta x}{\Delta t} + \frac{u_{n-2}^k - u_{n-1}^k}{\Lambda x} - \frac{x - (n-2) \Delta x}{\Delta t}
\end{aligned}
\right.
$$

From expressions (4.29) - (4.31), we deduce with the same kind of argument as for
(4.28) that

$$
(4.32) \quad \left| \frac{\partial \tilde{u}}{\partial x}(t) \right| \text{ is bounded independently of } h, \text{ and } t \in [0, T].
$$

We draw the same conclusions for $\tilde{u}_h$, and we deduce now that we may extract subsequences,
denoted by $\tilde{u}^h$ and $\tilde{u}^{h_0}$ such that
$$ \tilde{u}_h \rightarrow \bar{u}, \tilde{u} \rightarrow \tilde{u}_h \quad \text{in } L^\infty(0, T; V) \text{ weak } * $$

$$ \frac{\partial \bar{u}}{\partial t} \rightarrow \frac{\partial \tilde{u}}{\partial t}, \quad \frac{\partial \tilde{u}_h}{\partial t} \rightarrow \frac{\partial \tilde{u}}{\partial t} \quad \text{in } L^\infty(P, T, H) \text{ weak } * $$

$$ \bar{u}_h \rightarrow \bar{u}, \tilde{u}_h \rightarrow \tilde{u} \quad \text{in } C^0(Q_T) $$

To complete the proof, we have to write down variational inequalities satisfied by $\bar{u}_h$ and $\tilde{u}_h$. For $z$ in $V_h$ we call $\tilde{P}_h^k$ the element of $\tilde{V}_h^k$ which is equal to $z$ at the modes of $\tilde{G}_h^k$. We claim that, for any $\tilde{v}$ in $\tilde{V}_h^k$, the following relation holds:

$$ (4.33) \qquad a(\tilde{v}, y) = a(\tilde{v}, \tilde{P}_h^k y), \quad \forall y \in \tilde{V}_h^k. $$

To check this relation, we observe that if $j$ and $j+2$ belong to $G_h^k$, the relevant term on the left hand side of (4.33) is

$$ \begin{aligned} & \Delta x^{-1} \left[ \frac{V_{j+2} - V_j}{2} (Y_{j+1} - Y_j) + \frac{V_{j+2} - V_j}{2} (Y_{j+2} - Y_{j+1}) \right] = \\ & = \Delta x^{-1} \left( \frac{V_{j+2} - V_j}{2} \right) \left( \frac{Y_{j+2} - Y_j}{2} \right), \end{aligned} $$

and the relevant term on the right hand side of (4.33) is

$$ 2 \Delta x^{-1} \frac{V_{j+2} - V_j}{2} \frac{Y_{j+2} - Y_j}{2}, $$

which is clearly equal to the previous expression. If $j$ and $j+1$ belong to $\tilde{G}_h^k$, then the relevant terms are the same on the left and the right.

Let us now compute $s_h(\tilde{v}, y)$ in terms of $\tilde{s}^k(\tilde{v}, \tilde{P}_h^k y)$ and $s(\tilde{v}, \bar{P}_h^k y)$, for $\tilde{v}$ in $\tilde{V}_h^k$, $y$ in $V_h'$; we let

$$ \tilde{S}_h^k (v, y) = \sum_{j \in \tilde{G}_h^k} \tilde{v}(j \Delta x) y(j \Delta x) \Delta x (j - j') $$
and j' is the larger element of $\tilde{G}_h^k$, which is strictly smaller than j.

We have the following formula :

$$ (4.34) \quad \begin{aligned} & |s_h(\tilde{v}, y) - \frac{1}{2}\tilde{s}_h(\tilde{v}, \tilde{P}^k y)| \le \\ & \le C |y| \|\tilde{v}\| \sqrt{\Delta x} \end{aligned} $$

Let us prove this formula when, for instance, 1 and n = 1 belong to $\tilde{G}_h^k$; the other cases are of course analogous. If we let

$$ \tilde{v}(jh) := V_j, \quad jh \in \tilde{G}_h^k $$

$$ y(jh) = Y_j, $$

then

$$ s_h(\tilde{v}, y) = \Delta x \left[ V_1 Y_1 + \frac{V_1}{2} Y_2 + V_3 Y_3 + \dots \right] $$

$$ + \frac{V_{n-3} + V_{n-1}}{2} Y_{n-2} + V_{n-1} Y_{n-1} + V_n Y_n \] = $$

$$ = \Delta x \left[ V_1 \left(Y_1 + \frac{Y_1}{2}\right) + V_2 \left(\frac{Y_2}{2} + Y_3 + \frac{Y_3}{2}\right) + \dots \right] $$

$$ + V_{n-3} \left( \frac{Y_{n-4}}{2} + Y_{n-1} + \frac{Y_{n-2}}{2} \right) + V_{n-1} \left( \frac{Y_{n-4}}{2} + Y_{n-1} + \frac{Y_{n-2}}{2} \right) + V_n Y_n $$

and

$$ \tilde{s}_h(\tilde{v}, \tilde{P}_h^k y) = \Delta x \left[ V_1 Y_1 + 2 V_3 \dot{Y}_3 + \dots + 2 V_{n-1} Y_{n-1} + V_n Y_n \right] $$
$$S_h(\tilde{v}, \bar{P}_h^k y) = \Lambda x \left[ V_1 \frac{V_2}{2} + \frac{V_1 + V_3}{2} Y_2 + V_3 \frac{V_2 + V_4}{2} + \dots + \right. $$

$$+ \frac{V_{n-3} + V_{n-1}}{2} Y_{n-2} + V_{n-1} \frac{Y_{n-2} + Y_n}{2} + V_n Y_n] =$$

$$= \Lambda x \left[ V_1 Y_2 + V_3 (Y_2 + Y_4) + \dots + V_{n-1} \left(Y_{n-2} + \frac{Y_n}{2}\right) + V_n Y_n \right]$$

Therefore,

$$S_h(\tilde{v}, y) - \frac{1}{2} \tilde{S}_h^k(\tilde{v}, \tilde{P}_h^k y) - \frac{1}{2} S_h(\tilde{v}, \bar{P}_h^k y) = \Lambda x \left[ \frac{V_1 Y_1}{2} - \frac{V_{n-1} Y_n}{4} + V_n Y_n \right]$$

and

$$| \Lambda x \left[ \frac{V_1 Y_1}{2} - \frac{V_{n-1} Y_n}{4} + V_n Y_n \right] | \le$$

$$\le |y| \sqrt{\Delta x} \left( V_1^2 \Lambda x + V_{n-1}^2 \Lambda x + V_n^2 \Delta x \right)^{\frac{1}{2}}$$

$$\le C \|y\| |v| \sqrt{\Delta x}.$$

Similarly, let us estimate $a(\tilde{v}, y) - a(\tilde{v}, \tilde{P}_h^k y)$:

$$a(\tilde{v}, y) = \sum_{j=0}^{n-1} \frac{(V_j - V_i)(Y_i - Y_j)}{\Lambda x}$$

and of course, if $j$h does not belongs to $G_h^k$,

$$V_j = \frac{1}{2} (V_{j+1} + V_{j-1}).$$

In this case, the contribution from the interval with end points $(j-1)h$ and $(j+1)h$ is given by
$$ \Lambda_x \left( \frac{1}{2} \left( \frac{V_{j+1} - V_{j-1}}{\Lambda_x} \right) \left( \frac{\bar{V}_{j+1} - \bar{V}_{j-1}}{\Lambda_x} \right) + \frac{1}{2} \left( \frac{V_{j+1} - \bar{V}_{j-1}}{\Lambda_x} \right) \left( \frac{\bar{V}_{j+1} - \bar{V}_{j-1}}{\Lambda_x} \right) \right) \\ 2 \Lambda_x \left( \frac{V_{j+1} - \bar{V}_{j-1}}{2 \Lambda_x} \right) \left( \frac{\bar{V}_{j+1} - \bar{V}_{j-1}}{\Lambda_x} \right) $$

so that

$$ a(\tilde{v}, y) = a(\tilde{v}, \tilde{P}_h^k y), \quad \forall y \in V_h. $$

Let

$$ (4.35) \qquad u = \frac{1}{2} (\tilde{u} + \bar{u}) $$

let $v^n$ be as in (3.19), and denote

$$ (4.36) \qquad \phi^n = Q_h \left[ (v^n - u)(k+2)h \right] $$

Substitute in (4.5) at time $k$ by $u_h^{k+1} + \tilde{P}_h^k \phi^n$; with the help of (4.34), we obtain

$$ (4.37) \qquad \left\{ \begin{aligned} & \frac{1}{2} s_h^k \left( \frac{\bar{u}_h^{k+1} - 2\tilde{u}_h^k + \bar{u}_h^{k-1}}{\Delta t^2} , \tilde{P}_h^k \phi^n \right) \\ + & \frac{1}{2} s_h \left( \frac{\tilde{u}_h^{k+1} - 2\bar{u}_h^k + \tilde{u}_h^{k-1}}{\Delta t^2} , \tilde{P}_h^k \phi^n \right) \\ + & a (\tilde{u}_h^k, \tilde{P}_h^k \phi^n) \ge s_h (f_h^k, \tilde{P}_h^k \phi^n) + O(\sqrt{\Delta x}). \end{aligned} \right. $$

We have used in this substitution the following facts

$$ \tilde{v}_h^k = \bar{v}_h^{k+1}, \quad \tilde{P}_h^k = \bar{P}_h^{k+1}, \quad \tilde{s}_h^k = \bar{s}_h^{k+1}. $$
Similarly, substitute in (4.5) at time $k \ge 1$ v by $u_h^{k+2} + \tilde{P}_h^k \varphi^k$; we obtain

$$ (4.38) \quad \left\{ \begin{aligned} & \frac{1}{2} \tilde{S}_h^k \left( \frac{\tilde{u}_h^{k+2} - 2\bar{u}_h^{k+1} + \tilde{u}_h^k}{\Delta t^2} , \tilde{P}_h^k \varphi^k \right) + \\ & \quad + \frac{1}{2} S_h \left( \frac{\bar{u}_h^{k+2} - 2\tilde{u}_h^{k+1} + \bar{u}_h^k}{\Delta t^2} , \tilde{P}_h^k \varphi^k \right) + \\ & \quad + a (\bar{u}_h^{k+1}, \tilde{P}_h^k \varphi^k) \ge s_h (f_h^{k+1}, \tilde{P}_h^k \varphi^k) - O(\sqrt{\Delta x}). \end{aligned} \right. $$

We have a similar formula, centered at $k-1$:

$$ (4.39) \quad \left\{ \begin{aligned} & \frac{1}{2} \tilde{S}_h^k \left( \frac{\tilde{u}_h^k - 2\bar{u}_h^{k-1} + \tilde{u}_h^{k-2}}{\Delta t^2} , \tilde{P}_h^k \varphi^k \right) + \\ & \quad + \frac{1}{2} S_h \left( \frac{\bar{u}_h^k - 2\tilde{u}_h^{k-1} + \bar{u}_h^{k-2}}{\Delta t^2} , \tilde{P}_h^k \varphi^k \right) + \\ & \quad + a (\bar{u}_h^{k-1}, \tilde{P}_h^k \varphi^k) \ge s_h (f_h^{k-1}, \tilde{P}_h^k \varphi^k) - O(\sqrt{\Delta x}). \end{aligned} \right. $$

If we multiply (4.37) by 2, and add it to (4.38) and (4.39), we obtain eventually

$$ (4.40) \quad \left\{ \begin{aligned} & 2 \frac{\tilde{s}_h^k}{\Delta t} \left( \frac{\tilde{u}_h^{k+2} - 2\tilde{u}_h^k + \tilde{u}_h^{k-2}}{\Delta t^2} , \tilde{P}_h^k \varphi^k \right) + \\ & \quad + 2 S_h \left( \frac{\bar{u}_h^{k+2} - 2\bar{u}_h^k + \bar{u}_h^{k-2}}{(2\Delta x)^2} , \tilde{P}_h^k \varphi^k \right) + \\ & \quad + a \left( \bar{u}_h^{k+1} + 2\tilde{u}_h^k + \bar{u}_h^{k-1} , \tilde{P}_h^k \varphi^k \right) \ge \\ & \quad S_h \left( f_h^{k+1} + 2f_h^k + f_h^{k+1} , \tilde{P}_h^k \varphi^k \right) - O(\sqrt{\Delta x}). \end{aligned} \right. $$
By discrete integration, and techniques giving an estimate analogue to (4.16), we pass to the limit and obtain:

$$ - (u_t, v(0) - u_0) - \frac{1}{2} \int_0^T (\tilde{u}_t - \bar{u}_t, v_t - u_t) \, dt + \\ + \frac{1}{2} \int_0^T ((\tilde{u} + \bar{u}, v - u)) \, dt \ge \int_0^T (f, v - u) \, dt, $$

which proves that, thanks to the definition (4.35) of $u$, $u$ is the solution of (2.26)-(2.28). Working with the energy as in all previous proofs, one deduces in a straightforward way that the sequence $(\tilde{u}_h + \bar{u}_h)/2$ converges to $u$, and indeed that $\tilde{u}_h$ and $\bar{u}_h$ converge to $u$ in the strong topology of $L^2$. ■
V - NUMERICAL EXPERIMENTS.

We have taken L = 1, a contact at x = 1 and initial data such that the
solution of the linear problem would be a wave propagating to the right. We have
several cases, all in finite differences, corresponding to a situation when the initial
data is differentiable or not, or the scheme is characteristic or sub characteristic.

The smooth initial data is defined by

$$
u_0(x) = \begin{cases} -[(x-x_0)^2 - \alpha^2]^2 e^{-\alpha^4} & \text{on } [x_0 - \alpha, x_0 + \alpha] \\ 0 & \text{elsewhere} \end{cases}
$$

$$
u_0(x) = - du_0(x) / dx
$$

and, in our experiments,

$\alpha = 0.2$

$$
x_0 = 0.6 .
$$

The discretized initial data will be therefore

$$
\mathcal{U}_j^0 = u_0 (j \Lambda x)
$$

$$
\eta_j^1 = u_0 (j \Delta x - \Delta t)
$$

In the first set of figures (Fig. 2 to 11), we have taken a characteristic scheme, and
a space step of 1/50. The results look quite satisfactory, though a careful inspection
of the returning wave after reflexion on the side with a unilateral constraint reveals a
small hollow which does not exist in the exact solution (see Fig. 3). This small hollow
is due to the effect of the unilateral constraint which implies a change of boundary condi-
tion with the change in the sense of variation of the wave.

It will be more apparent with the piecewise affine data (see below).
In the second set of Figures (Fig. 12 to 19), with 51 discretization points and a subcharacteristic scheme with a CFL number of 0.8, the small hollow is not apparent, but the dispersive effect becomes large later. To decrease the magnitude of those dispersive effects, the spatial step is refined to 1/200, without changing the CFL number, and, as in the linear case, the results are substantially better; see Fig. 20 to 27.

The piecewise affine data are

$$u(x)_0 = -(1 - 5|x - 0.6|)^+$$

and a 51 points discretization and a characteristic scheme gives a "beak" in the middle of the returning wave, which is too large to make the simulation acceptable; a refinement in the time step gives a beak which is much smaller but still present; see Figures 28 to 41 and in particular, for the "beak", Fig. 32 to 40.

The same initial data, and a subcharacteristic scheme, with a CFL of 0.8 lead to the noticeable oscillations, where the beak was in the characteristic schemes, but the dispersive effect of this scheme improves the aspect of the solution with time; see Fig. 42 to 50.

Next set of figures (Fig 51 to 58) present the evolution in time of the solution with initial data

$$ \begin{array}{ll} u(x) = 0.0 & x < 0.5 \\ 50(x - 0.5) & 0.5 \le x < 0.51 \\ 0.5 & 0.51 \le x < 0.60 \\ 50(0.61 - x) & 0.60 \le x < 0.62 \\ -0.5 & 0.62 \le x < 0.72 \\ 50(x - 0.73) & 0.72 \le x < 0.73 \\ 0 & 0.73 \le x \le 0.8 \end{array} $$

We have tested the non-characteristic scheme on these data with steep derivatives; with a CFL number of 0.8 and 100 discretization points, the result has the normally expected oscillations, but is admissible.
Finally, in the last set of figures (Fig. 59 to 65), we display with a rather coarse discretization step (201 points) and a characteristic scheme, the evolution of the solution with data

$$
\begin{align*}
u(x) &= 0 && x < 0.25 \\
\cos(20\pi(x - 0.25)) && 0.25 \le x < 0.75 \\
0 && x > 0.75.
\end{align*}
$$

The humps are, in fact, returned one by one by the nonlinear constraint ; we have not
displayed all the iterations, to keep down their volume.

ACKNOWLEDGEMENT

We want to thank Mr Th. Dumont, who has been extremely helpful for computer graphics.
<figure><img src="image_4.png" /></figure>

<figure><img src="image_5.png" /></figure>
<figure><img src="image_6.png" /><figcaption>Figure 4</figcaption></figure>

ITERATION NO 41
DX/DT= 1.000

51 POINTS EN X

<figure><img src="image_7.png" /><figcaption>Figure 5</figcaption></figure>

ITERATION NO 61
DX/DT= 1.000

51 POINTS EN X
<figure><img src="image_8.png" /></figure>

<figure><img src="image_9.png" /></figure>
<figure><img src="image_10.png" /></figure>

<figure><img src="image_11.png" /></figure>
<figure><img src="image_12.png" /></figure>

<figure><img src="image_13.png" /></figure>
<figure><img src="image_14.png" /><figcaption>Figure 12</figcaption></figure>

<figure><img src="image_15.png" /><figcaption>Figure 13</figcaption></figure>
<figure><img src="image_16.png" /><figcaption>Figure 14</figcaption></figure>

ITERATION NO 101
OX/DT= 0.800
51 POINTS EN X

<figure><img src="image_17.png" /><figcaption>Figure 15</figcaption></figure>

ITERATION NO 151
OX/DT= 0.800
51 POINTS EN X
<figure><img src="image_18.png" /></figure>

<figure><img src="image_19.png" /></figure>
<figure><img src="image_20.png" /><figcaption>Figure 18</figcaption></figure>

ITERATION NO 301
DX/DT= 0.800
51 POINTS EN X

<figure><img src="image_21.png" /><figcaption>Figure 19</figcaption></figure>

ITERATION NO 351
DX/DT= 0.800
<figure><img src="image_22.png" /></figure>

<figure><img src="image_23.png" /></figure>

ITERATION NO 201
DX/DT= 0.800
201 POINTS EN X
<figure><img src="image_24.png" /><figcaption>Figure 22</figcaption></figure>

ITERATION NO 401
DX/DT= 0.800
201 POINTS EN X

<figure><img src="image_25.png" /><figcaption>Figure 23</figcaption></figure>

ITERATION NO 601
DX/DT= 0.800
201 POINTS EN X
<figure><img src="image_26.png" /><figcaption>Figure 24</figcaption></figure>

<figure><img src="image_27.png" /><figcaption>Figure 25</figcaption></figure>
<figure><img src="image_28.png" /></figure>

<figure><img src="image_29.png" /></figure>
<figure><img src="image_30.png" /></figure>

<figure><img src="image_31.png" /></figure>
<figure><img src="image_32.png" /></figure>

<figure><img src="image_33.png" /></figure>
<figure><img src="image_34.png" /><figcaption>Figure 32</figcaption></figure>

ITERATION NO 101
DX/DT= 1.000
201 POINTS EN X

<figure><img src="image_35.png" /><figcaption>Figure 33</figcaption></figure>

ITERATION NO 121
DX/DT= 1.000
201 POINTS EN X
<figure><img src="image_36.png" /></figure>

ITERATION NO 141
$\Delta X/\Delta T = 1.000$
201 POINTS EN X

<figure><img src="image_37.png" /></figure>

ITERATION NO 161
$\Delta X/\Delta T = 1.000$
201 POINTS EN X
<figure><img src="image_38.png" /><figcaption>Figure 36</figcaption></figure>

<figure><img src="image_39.png" /><figcaption>Figure 37</figcaption></figure>
<figure><img src="image_40.png" /></figure>

<figure><img src="image_41.png" /></figure>
<figure><img src="image_42.png" /></figure>

ITERATION NO 261
DX/DT= 1.000
201 POINTS EN X

<figure><img src="image_43.png" /></figure>

ITERATION NO 281
DX/DT= 1.000
201 POINTS EN X
<figure><img src="image_44.png" /></figure>

<figure><img src="image_45.png" /></figure>
<figure><img src="image_46.png" /><figcaption>Figure 44</figcaption></figure>

ITERATION NO 129
DX/DT= 0.800
201 POINTS EN X

<figure><img src="image_47.png" /><figcaption>Figure 45</figcaption></figure>

ITERATION NO 193
DX/DT= 0.800
201 POINTS EN X
<figure><img src="image_48.png" /><figcaption>Figure 46</figcaption></figure>

ITERATION NO 257
DX/DT= 0.800
201 POINTS EN X

<figure><img src="image_49.png" /><figcaption>Figure 47</figcaption></figure>

ITERATION NO 321
DX/DT= 0.800
201 POINTS EN X
<figure><img src="image_50.png" /></figure>

<figure><img src="image_51.png" /></figure>
<figure><img src="image_52.png" /></figure>

IITERATION NO 513

DX/DT: 0.800

201 POINTS EN X
<figure><img src="image_53.png" /><figcaption>Figure 51</figcaption></figure>

<figure><img src="image_54.png" /><figcaption>Figure 52</figcaption></figure>
<figure><img src="image_55.png" /></figure>

<figure><img src="image_56.png" /></figure>
<figure><img src="image_57.png" /></figure>

<figure><img src="image_58.png" /></figure>
<figure><img src="image_59.png" /></figure>

<figure><img src="image_60.png" /></figure>
<figure><img src="image_61.png" /><figcaption>Figure 59</figcaption></figure>

<figure><img src="image_62.png" /><figcaption>Figure 60</figcaption></figure>
<figure><img src="image_63.png" /></figure>

<figure><img src="image_64.png" /></figure>
<figure><img src="image_65.png" /><figcaption>Figure 63</figcaption></figure>

<figure><img src="image_66.png" /><figcaption>Figure 64</figcaption></figure>
<figure><img src="image_67.png" /></figure>

IITERATION NO4801
DX/DT= 1.000
201 POINTS EN X
REFERENCES

[1] L. Amerio, Su un problema di vincoli unilaterali per l'equazione non omogenea della corda vibrante, IAC (Istituto per le applicazioni del calcolo "Mauro Picone") Pubbl. Ser. D 109 (1976) 3 - 11.

[2] H. Brezis, *Opérateurs maximaux monotones et semi-groupes de contraction dans les espaces de Hilbert*, North Holland, Amsterdam-London-New York, 1973.

[3] C. Citrini, The energy theorem in the impact of a string vibrating against a point shaped obstacle - Rend. Acc. Naz. Lincei 62 (1977) 143 - 149.

[4] C. Citrini, Risultati tipici sul problema della corda vibrante con ostacolo puntiforme, IAC (Istituto per le applicazioni del calcolo "Mauro Picone") Pubbl., Ser III 134 (1978) 1 - 24.

[5] G. Lebeau, M. Schatzman, A wawe problem in a half-space with a unilateral condition at the boundary, J. Differential Equations, 53 (1984) 309 - 361.

[6] J.L. Lions, Quelques méthodes de résolution de problèmes aux limites non-linéaires, Dunod et Gauthier-Villars, Paris, 1969.

[7] R.S. Phillips, The adjoint semi-group, Pacific J. Math. 5 (1955) 269 - 283.

[8] R.T. Rockafellar, Integrals which are convex functionals, II, Pacific J. Math. 39 (1971) 439 - 469.

[9] M. Schatzman, A class of nonlinear differential equations of second order in time, Nonlinear Analysis, Theory, methods and Applications, 2 (1978) 355 - 373.

[10] M. Schatzman, Un problème hyperbolique du 2ème ordre avec contrainte unilatérale : la corde vibrante avec obstacle ponctuel, J. Differential Equations, 36 (1980) 295 - 334.

[11] M. Crouzeix, A.L. Mignot, Analyse numérique des équations différentielles, Masson, Paris 1984.