# A class of nonlinear differential equations of second order in time

Michelle Schatzman

► To cite this version:

Michelle Schatzman. A class of nonlinear differential equations of second order in time. Nonlinear Analysis: Theory, Methods and Applications, 1978, 2 (3), pp.355-373. 10.1016/0362-546X(78)90022-6. hal-01294058

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<figure><img src="image_1.png" /></figure>
# A CLASS OF NONLINEAR DIFFERENTIAL EQUATIONS
OF SECOND ORDER IN TIME

MICHELLE SCHATZMAN

Analyse Numerique t55-65, Universite P. et M. Curie, 4pl Jussieu, 75230 Paris Cedex 05, France

*Key words:* Existence and uniqueness, nonlinear equations of second order, monotone operators.

## 1. INTRODUCTION

IN WHAT follows, $\phi$ will denote a lower semicontinuous convex proper function from $\mathbb{R}^n = H$ to $\mathbb{R} \cup \{+\infty\}$. Here proper means that $\phi \not\equiv +\infty$. The effective domain of $\phi$ is the set $\text{dom } \phi = \{x \in \mathbb{R}^n/\phi(x) < +\infty\}$. We shall suppose that the interior of $\text{dom } \phi$ in $\mathbb{R}^n$ is not empty, and $\phi \ge 0$. These two assumptions do not restrict the generality. The scalar product in $H$ is denoted by $(x, y)$.

Let us recall that the subdifferential of $\phi$ is a (multivalued) operator defined on

$$D(\partial\phi) = \{x \in H : \exists z \in H \text{ such that } \phi(x + y) - \phi(x) \ge (z, y) \text{ for all } y \in H\}$$

with values

$$\partial\phi(x) = \{z \in H : \phi(x + y) - \phi(x) \ge (z, y), \forall y \in H\}.$$

The set $\partial\phi(x)$ is closed convex in $H$ and $\partial\phi$ is a cyclically monotone operator, i.e. for all $n \in \mathbb{N}$, for all $n$-tuple $(x_1, \dots, x_n)$ of elements of $D(\partial\phi)$, for all $y_i \in \partial\phi(x_i)$ ($1 \le i \le n$)

$$(y_n, x_n - x_{n-1}) + \dots + (y_2, x_2 - x_1) + (y_1, x_1 - x_n) \ge 0.$$

We define $J_\lambda = (I + \lambda\partial\phi)^{-1}$ on $H$; $J_\lambda$ is a contraction, and $\partial\phi_\lambda = (I + J_\lambda)/\lambda$ is the Yosida approximation to $\partial\phi$; $\partial\phi_\lambda$ is Lipschitzian, with Lipschitz constant $1/\lambda$ and $\partial\phi_\lambda(x) \in \partial\phi(J_\lambda x)$. If $\phi_\lambda(x) = \inf\{|x - y|^2/(2\lambda) + (\phi/2)(y)\}, \partial\phi_\lambda$ is the subdifferential of $\phi_\lambda$. Explicitly, $\phi_\lambda(x) = \phi(J_\lambda x) + |x + J_\lambda x|^2/(2\lambda)$ and $\phi_\lambda(x) \le \phi(x)$. Moreover, $\lim_{\lambda \to 0} \phi_\lambda(x) - \phi(x)$ if $x \in H$. The general references for these results are refs [1, 2].

Before we proceed to the existence theorem, it is of some interest to give an explicit example. It will allow the reader to notice the difficulties of the problem.

Take $H = \mathbb{R}$, $K = \mathbb{R}^+$, $\phi = \psi$ the indicator function of $K$, i.e.

$$\phi(x) = +\infty \quad \text{if } x \notin K$$

$$\phi(x) = 0 \quad \text{if } x \in K$$

Then

$$\partial\phi(x) = \{0\} \quad \text{if } x > 0$$

$$\partial\phi(x) = (-\infty; 0] \quad \text{if } x = 0$$

$$\partial\phi(x) = \emptyset \quad \text{if } x < 0.$$
Assume $u_0 > 0$; we shall seek a solution $u$, which is locally Lipschitzian in $t$. The interesting case occurs when $u_1 < 0$. If $t < u_0/u_1$, the only solution of

$$ \frac{d^2 u}{dt^2} + \partial \psi_k(u) \ge 0 $$

is $u(t) = x_0 + tu_1$. We must have $u(t) \ge 0$, for all $t \ge 0$. It is easy to check that any $u$ of the form

$$ u(t) = 0 \quad \text{for} \quad t_0 = -u_0/u_1 \le t \le t_1 $$

$$ u(t) = v(t - t_1) \quad \text{for} \quad t \ge t_1 $$

with $v > 0$ and $-u_0/u_1 \le t_1 \le \infty$ is a solution. (see the figure below). If we assume that the energy is conserved, then necessarily $u^0(t) = -u_1(t + u_0|u_1)$ for $t \ge -u_0/u_1$. There is an

<figure><img src="image_2.png" /><figcaption>Fig. 1.</figcaption></figure>

“optical reflection” in $-u_0/u_1$. The Yosida approximation gives, in the limit, the solution which conserves the energy. In fact, the approximating equation is

$$ \frac{d^2 u_{\lambda}}{dt^2} - \frac{u_{\lambda}^{-}}{\lambda} = 0, \text{ where } r^{-} = -\min(r, 0); \text{ its solution is} $$

$$ u_{\lambda}(t) = u_0 + tu_1 $$

$$ 0 \le t \le -u_0/u_1 = t_0 $$

$$ u_{\lambda}(t) = \sqrt{\lambda} u_1 \sin \left[ \frac{t+t_0}{\sqrt{\lambda}} \right] $$

$$ t_0 \le t \le t_0 + \pi \sqrt{\lambda} $$

$$ u_{\lambda}(t) = -u_1(t - \pi \sqrt{\lambda} - t_0) $$

$$ t_0 + \pi \sqrt{\lambda} \le t. $$

Clearly, the $u_{\lambda}$ converge to the energy conserving solution, $u^0$.

## 2. THE EXISTENCE THEOREM

*Definition 1.* If $f \in L^2(0, T; H)$ ($T$ finite) and $u_0 \in \text{dom } \phi$, $u_1 \in \tilde{H}$, we say that $u$ is an energy-conserving solution of the problem (P)
$$
(P) \qquad (d^2u/dt^2) + \partial\phi(u) \ni f, \quad \text{"initial conditions"}
$$

when it satisfies the following requirements:

$$
(1) u \in W^{1, \infty}((0, T); H)
$$

$$
(2) u(t) \in \operatorname{dom} \phi, \forall t \in [0, T]
$$

(3) there exists a bounded measure $\mu$ with values in $H$, such that
$(d^2u/dt^2) + \mu = f$ in the sense of distributions,

and

(4) for any continuous $v$, with values in $H$, such that $\phi(v) \in L^1(0, T)$, we have:

$$
\int_0^T (\phi(v) - \phi(u)) \, dt \geq \langle \mu, v - u \rangle
$$

(5) du/dt has left and right limits for any $t \in [0, T]$ (with the necessary modifications at 0 and T)

(6) energy is conserved:

$$
\left| \left| \frac{d^+ u}{dt}(t) \right|^2 + \phi(u(t)) \right| = \left| \left| \frac{d^- u}{dt}(t) \right|^2 + \phi(u(t)) \right| = |u_1|^2 + \phi(u_0) + \int_0^t (\dot{u}(s), f(s)) \, ds
$$

almost everywhere on $[0, T]$; here $\dot{u}$ stands for $du/dt$

(7) the initial conditions are satisfied in the following sense:

$u(0) = u_0$

and if $K_0$ is the closure of $\text{dom } \phi, \psi_{K_0}$ the indicator function of $K_0$:

$$
-u_1 + \frac{d^+ u}{dt}(0) + \partial\psi_{K_0}(u_0) \ni 0.
$$

THEOREM 1. (a) For any $f \in L^2(0, T; H)$, $u_0 \in \text{dom } \phi$ and $u_1 \in H$, the problem (P) has an energy-
conserving solution in the sense of Definition 1. This solution is obtained as the strong limit in
$H^1((0, T); H)$ and the weak * limit in $W^{1, \infty}((0, T); H)$ of a subsequence of the sequence of solu-
tions of $(P_\lambda)$:

$$
(P_\lambda) \qquad \frac{d^2 u_\lambda}{dt^2} + \partial \phi_\lambda(u_\lambda) = 0, \quad u_\lambda(0) = u_0, \quad \frac{d u_\lambda}{dt}(0) = u_1.
$$

(b) Moreover, if $\phi$ is Lipschitzian in a neighbourhood of $u_0$ with respect to $K_0$, and if $-u_1$
belongs to $C$, the tangent cone to $K_0$ at $u_0$ (i.e. $C = \bigcup_{\tau>0} \tau(K_0 - u_0)$), then

$$
(8) \qquad \frac{d^+ u}{dt}(0) = 2P_c u_1 - u_1
$$

where $P_c$ is the projection onto $C$.

Proof. Part (a)

1. Preliminary estimates

We have the energy equality for $(P_\lambda)$

$$
(9) \qquad \frac{1}{2}|\dot{u}_{\lambda}(t)|^2 + \phi_{\lambda}(u_{\lambda}(t)) = \frac{1}{2}|u_1|^2 + \phi_{\lambda}(u_0) + \int_0^t (\dot{u}_{\lambda}(s), f(s)) \, ds
$$
from which follows the inequality

$$
\begin{aligned}
\frac{1}{2} |\dot{u}_{\lambda}(t)|^2 &\le \frac{1}{2} |u_1|^2 + \phi(u_0) + |f|_{L^2(0, T; H)} \left( \int_0^t |\dot{u}_{\lambda}(s)|^2 ds \right)^{\frac{1}{2}} \\
&\le \frac{1}{2} |u_1|^2 + \phi(u_0) + \frac{1}{2} |f|_{L^2(0, T; H)} \left( 1 + \int_0^t |\dot{u}_{\lambda}(s)|^2 ds \right)
\end{aligned}
$$

by the relation $x \le (1 + x^2)/2$.

Gronwall's lemma implies:

$$
\phi_{\lambda}(u_{\lambda}(t)) + \left| \frac{1}{2} \dot{u}_{\lambda}(t) \right|^{2} \le \left( \frac{1}{2} |u_{1}|^{2} + \phi(u_{0}) + \frac{1}{2} |f|_{L^{2}(0, T; H)} \right) \exp \left( T |f|_{L^{2}(0, T; H)} \right)
$$

Set

$$
(10) \qquad E^2/2 = \left( \frac{1}{2} |u_1|^2 + \phi(u_0) + \frac{1}{2} |f|_{L^2(0, T; H)} \right) \exp(T|f|_{L^2(0, T; H)})
$$

We may therefore extract a subsequence, still denoted by $u_\lambda$, such that

$u_\lambda \to u$ in $C^0([0, T]; H)$ strong

$\frac{du_\lambda}{dt} \to \frac{du}{dt}$ in $L^\infty(0, T; H)$ weak*

$\phi_\lambda(u_\lambda) \to \chi$ in $L^\infty(0, T)$ weak*.

As $\phi_\lambda(u_\lambda) \le E, |u_\lambda - J_\lambda u_\lambda| \le \sqrt{2\lambda E}$ and $\phi(J_\lambda u_\lambda) \le E$. We have

$$
E \geq \lim_{\lambda \to 0} \phi(J_\lambda u_\lambda(t)) \geq \phi(u(t))
$$

and $u(t) \in \text{dom } \phi$, for all $t$ in $[0, T]$. Hence we obtain (1) and (2).

2. Main estimate

Let $a$ be the center of a closed ball of radius $\rho$, contained in the interior of dom $\phi$. By general theorems, we may suppose that $\phi$ is bounded on $a + B_\rho$ by a constant $C$. For an arbitrary continuous $z$ such that $|z(t)| \le 1$ for all $t$ in $[0, T]$, we may write

$$
(\partial\phi_{\lambda}(u(t)), a + z(t) - u(t)) \le \phi_{\lambda}(a + \rho z(t)) - \phi_{\lambda}(u_{\lambda}(t)).
$$

Integrating from 0 to T, we have

$$
\begin{align*}
\rho \int_0^T (\partial \phi_\lambda(u_\lambda(t)), z(t)) \, dt &\le CT + \int_0^T \left( \frac{d^2 u_\lambda}{dt^2} - f, a - u_\lambda \right) \, dt \\
&\le CT + \left( \frac{du_\lambda}{dt}, a - u_\lambda \right)_0^T + \int_0^T \left| \frac{du_\lambda}{dt} \right|^2 \, dt + \int_0^T (f, a - u_\lambda) \, dt \\
&\le CT + E(2|a| + 2|u_0| + TE) + TE^2 + |f|_{L^2(0,T;H)} \sqrt{T(|a| + |u_0| + TE)}.
\end{align*}
$$

If we choose

$$
z(t) = \frac{1}{\rho} \frac{\partial \phi_{\lambda}(u_{\lambda}(t))}{\partial \phi_{\lambda}(u_{\lambda}(t))}
$$

we obtain the estimate
$$
(11) \qquad \int_0^T |\partial\phi_\lambda(u_\lambda(t))| \, dt \le C' \text{ independent of } \lambda.
$$

We may therefore extract a new subsequence, still denoted by $u_\lambda$, such that $\partial\phi_\lambda(u_\lambda) \to \mu$ vaguely in $M^1([0, T]; H)$, the set of bounded measures on $[0, T]$ with values in $H$. In the sense of distributions, $d^2u/dt^2 + \mu = f$, which is precisely (3).

If $v$ is an arbitrary element of $C^0([0, T]; H)$,

$$
\lim_{\lambda \to 0} \phi_{\lambda}(v(t)) = \phi(v(t)).
$$

Let $v$ be such that $\phi(v) \in L^1(0, T)$. Then $\phi(v_\lambda)$ is also in $L^1(0, T)$. We have

$$
(12) \quad \int_0^T (\phi_\lambda(v(t)) - \phi_\lambda(u_\lambda(t))) \, dt \ge \int_0^T (\partial\phi_\lambda(u_\lambda(t)), v(t) - u_\lambda(t)) \, dt.
$$

It is clear that the right-hand side tends to

$$
\langle \mu, v - u \rangle
$$

as $\lambda$ goes to infinity.

We have

$$
\lim \int_0^T \phi_\lambda(u_\lambda(t)) \ge \int_0^T \phi_\lambda(u_\lambda(t)) \, dt \le \int_0^T \phi(u(t)) \, dt.
$$

Hence, taking the upper limit in (12), we get

$$
\int_0^T (\phi(v(t)) - \phi(u(t))) \, dt \ge \langle \mu, v - u \rangle, \text{i.e. (4)}
$$

Interpretation of $\mu$.

Let $d\mu = g \cdot dt + d\mu_s$ be the decomposition of $\mu$ with respect to the Lebesgue measure on $[0, T]$.
Then Corollary 5.A of ref. [5], gives us the following results:

(13) $g(t) \in \partial\phi(u(t))$ almost everywhere (with respect to the Lebesgue measure) on $[0, T]$.

Let $N(t)$ be the normal cone to $K_0 = \overline{\text{dom } \phi}$ at $u(t)$. Let $\mu_s = h \cdot |\mu_s|$ with $h$ a $\mu_s$-integrable function.
Then

(14) $h(t) \in N(t) \quad |\mu_s| -$ almost everywhere on $[0, T]$.

We shall often say in what follows that $\mu$ is the measure associated with the solution $u$ of (P).

3. The properties of du/dt

We know that the injection $M^1(0, T) \subsetneq H^{-1}(0, T)$ is compact. As $\mu_\lambda$ converges vaguely to $\mu$ in $M^1(0, T; H) \cong (M^1(0, T))^N$, $\mu_\lambda$ converges strongly to $\mu$ in $H^{-1}(0, T; H) \cong (H^{-1}(0, T))^N$, and therefore

$$
\frac{du}{dt} \rightarrow \frac{du}{dt} \text{ strongly in } L^2(0, T; H).
$$

As $|du_\lambda/dt| \le E$, for all $t$ and $\lambda$, $(du_\lambda/dt) \to (du/dt)$ in $L^2(0, T; H)$, for all $p \in [1, \infty)$. On the other hand, as $d^2u/dt^2$ is a measure, $du/dt$ has right and left limits in every point of $(0, T)$, a right limit in $0$, and a left limit in $T$, and we can write
$$ \frac{d^{-}u}{dt} = u_{1} + \int_{0}^{T} f(s) \, ds - \mu([0, T]) $$

$$ \frac{d^{+}u}{dt} = u_{1} + \int_{0}^{T} f(s) \, ds - \mu([0, T]). $$

### 4. Energy conservation
It is clear that

$$ \lim_{\lambda \to 0} \int_{0}^{T} \frac{|u_{\lambda} - J_{\lambda}u_{\lambda}|^{2}}{\lambda} dt = 0, $$

because $u_{\lambda} - J_{\lambda}u_{\lambda} \to 0$ in $C^{0}([0, T]; H)$ and $(u_{\lambda} - J_{\lambda}u_{\lambda})/\lambda \to \mu$ in $M^{1}([0, T]; H)$. On the other hand, $|u_{\lambda} - J_{\lambda}u_{\lambda}|^{2}/\lambda$ is bounded in $L^{\infty}([0, T])$. Thus, $|u_{\lambda} - J_{\lambda}u_{\lambda}|^{2}/\lambda$ converges to 0 in $L^{\infty}([0, T])$ weak*. As $\phi_{\lambda}(u_{\lambda})$ converges to a certain $\chi$, $\phi(J_{\lambda}u_{\lambda}) - \phi_{\lambda}(u_{\lambda}) - |u_{\lambda} + J_{\lambda}u_{\lambda}|^{2}/(2\lambda)$ has the same limit in $L^{\infty}([0, T])$ weak*. We have the inequality

$$ \int_{0}^{T} (\phi(u) - \phi(J_{\lambda}u_{\lambda})) dt \geq \langle \mu_{\lambda}, u - J_{\lambda}u_{\lambda} \rangle \quad \text{as } \mu_{\lambda}(t) \in \partial\phi(J_{\lambda}u_{\lambda}(t)) \text{ for } t \in [0, T]. $$

Passing to the limit:

$$ \int_{0}^{T} (\phi(u) - \chi) \, dt \geq 0. $$

But:

$$ \lim_{\lambda \to 0} \phi(J_{\lambda}u_{\lambda}) \geq \phi(u). $$

This implies $\chi \geq \phi(u)$, thus proving that $\chi = \phi(u)$. If $\mu(\{t_0\}) = 0$, we see from (9) that

$$ \lim_{\lambda \to 0} \phi_{\lambda}(u_{\lambda}(t_0)) = \lim_{\lambda \to 0} \phi(u_{\lambda}(t_0)) = \frac{1}{2}|u_1|^2 + \phi(u_0) + \int_0^t (\dot{u}(s)f(s)) \, ds; $$

hence:

$$ \frac{1}{2}|\dot{u}(t)|^2 + \phi(u(t)) = \frac{1}{2}|u_1|^2 + \phi(u_0) + \int_0^t (\dot{u}(s), f(s)) \, ds, \text{ almost everywhere on } [0, T]. $$

### 5. Initial conditions
It is quite clear that $u(0) = u_0$.

To finish the proof, we need only consider the interpretation (13), (14) of $\mu$. In fact, $N(t) = \partial\psi_{K_0}(u(t))$. If $d^+u/dt(0) \neq u_1$, then $\mu_s = (-u_1 + d^+u/dt(0))\delta_0 + \tilde{\mu}_{s'}$ and the conclusion of Part (a) of Theorem 1 holds.

#### Part (b)

The case when $u_0 \in \text{int } K_0$ is quite simple. So we make the hypothesis $u_0 \in \partial K_0$. The idea of the proof is to compare $u_{\lambda}$ with the solution $v_{\lambda}$ of

$$ (15) \qquad \begin{cases} \frac{d^2 v_{\lambda}}{dt^2} + (\partial \psi_C)_\lambda (v_\lambda - u_0) = 0 \\ v_\lambda(0) = u_0 \\ \frac{dv_\lambda}{dt} = u_1 \end{cases} $$
Notice that $\partial\psi_{C\lambda}(x) = (x - P_C x)/\lambda$;

$v_\lambda$ is given explicitly by

$$v_{\lambda}(t) = u_{0} + tP_{C}u_{1} + \sqrt{\lambda}\sin(t/\sqrt{\lambda})(u_{1} - P_{C}u_{1}) \quad \text{if } 0 \le t \le \pi\sqrt{\lambda}$$

$$v_{\lambda}(t) = u_{0} + tP_{C}u_{1} + (t - \pi\sqrt{\lambda})(u_{1} - P_{C}u_{1}) \quad \text{if } t \ge \pi\sqrt{\lambda}$$

We can check these formulas using the fact that the decomposition of an arbitrary element of $H$ into the sum of its projections onto $C$ and $C^{\perp} = N$ is unique. This result is given in ref. [3], Lemma 2.2. Knowing that $(\partial\psi_c)_\lambda$ is Lipschitzian, we can see that $v_\lambda$ is the solution of (15). Let us show now that $\partial\phi_\lambda$ is locally near $(\partial\psi_c)_\lambda$ in an adequate sense.

**PROPOSITION 1.** If $\phi$ is Lipschitzian in a neighbourhood of $u_0$, there exists positive $\rho$ and $\lambda_0$ such that, for $|x - u_0| \le \rho$ and $\lambda \le \lambda_0$:

$$(16) \qquad (\partial\psi)_{\lambda}(x) - \partial\phi_{\lambda}(x) = h(x - u_0)/\lambda + A_{\lambda}(x)$$

where: $h(x - u_0) = P_{K_0}(x) - P_C(x - u_0)$;

$|h(x)| \le |x|\epsilon(|x|)$ and $\epsilon(r)$ decreases to zero as $r$ goes to zero;

$h(x) = 0$ on $K_0 - u_0$

$$|A_{\lambda}(x)| \le k$$

*Proof.* 1. Let $\phi$ be Lipschitzian over $\text{dom } \phi \cap \overline{B_{\sigma}(u_0)}$ with Lipschitz constant $k$. Then $\text{dom } \phi \cap \overline{B_{\sigma}(u_0)}$ is closed: in fact, if $(x_n)_n$, is a sequence of points of $\text{dom } \phi \cap \overline{B_{\sigma}(u_0)}$ converging to $x_\infty$, we have

$$\phi(x) \le \lim_{n \to \infty} \phi(x_n) \le \lim_{n \to \infty} (k|x_n - x_p| + \phi(x_p)) = \phi(x_p) + |x - x_p|,$$

which shows that $x \in \text{dom } \phi$.

2. Lipschitzian extension of $\phi|_{\overline{B_{\sigma}(u_0)}}$.

Denote by $\tilde{\phi}$ the convex function equal to $\phi$ in $\text{dom } \phi \cup \overline{B_{\sigma}(u_0)}$ and $+\infty$ elsewhere. Obviously $\tilde{\phi}$ is lower semicontinuous.

Define a convex function $\chi$ by

$$\chi(x) = \sup\{\tilde{\phi}(y) + (z, x - y): y \in \text{int } \text{dom } \tilde{\phi} \text{ and } z \in \partial\tilde{\phi}(y)\}.$$

In fact $\chi$ is a Lipschitzian extension of $\tilde{\phi}|_{\text{dom } \tilde{\phi}}: \chi(x)$ is never infinite, thanks to the relation

$$\chi(x) \ge \tilde{\phi}(y) + k|x - y|$$

(clearly $\partial\tilde{\phi}$ is bounded by $k$ on $\text{int } \text{dom } \tilde{\phi}$).

If $x \in \text{dom } \tilde{\phi}$, $\chi(x) \ge \tilde{\phi}(x)$. On the other hand, given $\varepsilon > 0$, there exists a $y_\varepsilon \in \text{int } \text{dom } \tilde{\phi}$ and a $z_\varepsilon \in \partial\tilde{\phi}(y_\varepsilon)$ such that

$$\chi(x) \le \tilde{\phi}(y_\varepsilon) + (z_\varepsilon, x - y_\varepsilon) + \varepsilon.$$

Therefore

$$\chi(x) \le \tilde{\phi}(x) + \varepsilon.$$

As $\varepsilon$ is arbitrary, we obtain $\chi(x) = \tilde{\phi}(x)$. We can extend this equality to all of the domain of $\tilde{\phi}$, by continuity. Let us show that $\chi$ is Lipschitzian: let $x$ and $x'$ be given in $H$; we have

$$\chi(x') - \chi(x) \le \tilde{\phi}(y'_e) + (z'_e, x' - y'_e) + \varepsilon - \tilde{\phi}(y'_e, x - y'_e)$$

$$= (z'_e, x' - x) + \varepsilon \le k|x' - x| + \varepsilon,$$
if $\epsilon$ is arbitrary and $y'_s$ and $z'_s$ are chosen adequately. We can bound from below $\chi(x') - \chi(x)$ by the same type of argument, and thus we have shown that $|\chi(x') - \chi(x)| \le k|x' + x|$.

3. Decomposition of $\tilde{\phi}$.

Denote by $\psi$ the indicator function of dom $\tilde{\phi} = \bar{K}$.

Obviously $\tilde{\phi} = \psi + \chi$; then by Theorem 23.8 of ref. [2], $\partial\tilde{\phi}(x) = \partial\chi(x) + \partial\psi(x)$ for all $x$ in $\bar{K}$.
We have

$$ \partial\tilde{\phi}_{\lambda}(x) = (x - (I + \lambda\partial\tilde{\phi})^{-1}x)/\lambda. $$

Let $(I + \lambda\partial\tilde{\phi})^{-1}x = z_{\lambda}$:

$$ z_{\lambda} + \lambda\partial\psi(z_{\lambda}) + \lambda g_{\lambda} \ni x $$

where $g_{\lambda} \in \partial\chi(z_{\lambda})$ and is thus bounded by $k$;

$$ z_{\lambda} = (I + \lambda\partial\psi)^{-1}(x - \lambda g_{\lambda}). $$

$$ \begin{aligned} |\partial\tilde{\phi}_{\lambda}(x) - \partial\psi_{\lambda}(x)| &= |[x - (I + \lambda\partial\psi)^{-1}(x - \lambda g_{\lambda}) - x + (I + \lambda\partial\psi)^{-1}x]|/\lambda \\ &= 1/\lambda |(I + \lambda\partial\psi)^{-1}(x - \lambda g_{\lambda}) - (I + \lambda\partial\psi)^{-1}x| \le |g_{\lambda}| \le k. \end{aligned} $$

Lemma 4.6. of Zarantonello [3] allows us to write: $P_{\mathbb{R}}(x + u_0) = u_0 + P_{\mathbb{C}}x + h(x)$ with

$$ |h(x)| \le |x|\varepsilon(|x|) \quad \text{and} \quad h(x) = 0 \text{ if } x \in \bar{K} - u_0. $$

Then

$$ |\partial\psi_{\lambda}(x) - (\partial\psi_{\mathbb{C}})_{\lambda}(x - u_0)| = |P_{\mathbb{C}}(x - u_0) - P_{\mathbb{R}}(x - u_0)|/\lambda \le \varepsilon(|x - u_0|)|x - u_0|. $$

4. Comparison of $\partial\phi_{\lambda}$ and $\partial\tilde{\phi}_{\lambda}$

The sequence $(I + \lambda\partial\phi)^{-1}$ converges to $P_{K_0}x$ when $\lambda$ goes to zero; but $(I + \lambda\partial\phi)^{-1}$ is a contraction; this convergence is therefore uniform on compact sets. So, there exists a function $j$ of $\rho$ and $\lambda$, decreasing to zero as $\lambda \to 0$, such that $|(I + \lambda\partial\phi)^{-1}x - P_{K_0}x| \le j(\lambda, \rho)$ on the ball $B_{\rho}(u_0)$, and in an analogous fashion

$$ |(I + \lambda\partial\tilde{\phi})^{-1}x - P_{\mathbb{R}}x| \le j(\lambda, \rho) \quad \text{on the ball } B_{\rho}(u_0). $$

$$ \text{Clearly, } P_{K_0}(B_{\rho}(u_0)) \subset B_{\rho}(u_0) \quad \forall \rho > 0 $$

$$ P_{\bar{K}}(B_{\rho}(u_0)) \subset B_{\rho}(u_0) \quad \forall \rho > 0. $$

Take $\rho = \sigma/2$, and choose $\lambda_0$ so small that

$$ \max(j(\lambda_0, \sigma/2), j(\lambda_0, \sigma/2)) < \sigma/2. $$

Then $|(I + \lambda\partial\phi)^{-1}x - u_0| < \sigma$, and $|(I + \lambda\partial\tilde{\phi})^{-1}x - u_0| < \sigma$ if $|x| < \sigma/2$. Let $y = (I + \lambda\partial\phi)^{-1}x$ and $\tilde{y} = (I + \lambda\partial\tilde{\phi})^{-1}x$. We know that $\phi|_{B_{\sigma}(u_0)} = \tilde{\phi}|_{B_{\sigma}(u_0)}$.

Therefore

$$ (1/\lambda)(x - y, v) + \phi(y + v) - \phi(y) \ge 0 \quad \forall v \in H $$

$$ (1/\lambda)(x - \tilde{y}, \tilde{v}) + \tilde{\phi}(\tilde{y} + \tilde{v}) - \tilde{\phi}(\tilde{y}) \ge 0 \quad \forall \tilde{v} \in H. $$

If we add these two inequalities, taking $v = \tilde{y} - y$, and $\tilde{v} = -v$, we obtain $\tilde{y} = y$. This achieves the proof of Proposition 1.
LEMMA 1. $u_\lambda$ and $v_\lambda$ satisfy the inequalities:

$$
(17) \qquad |u_{\lambda}(t) - v_{\lambda}(t)| \le ch \frac{t}{\sqrt{\lambda}} \left( k \frac{t^2}{2} + E \frac{t^3}{3!\lambda} \varepsilon(Et) \right)
$$

$$
(18) \qquad |\dot{u}_{\lambda}(t) - \dot{v}_{\lambda}(t)| \le kt + \frac{Et^2}{2\lambda}\varepsilon(Et) + \frac{1}{\lambda}ch\frac{t}{\sqrt{\lambda}}\left(k\frac{t^3}{3!} + \frac{Et^4}{4!\lambda}\varepsilon(Et)\right).
$$

*Proof*. Define a transformation $T_\lambda$ on $C^o([0, T]; H)$ by

$$
(T_{\lambda}w)(t) = u_{0} + tu_{1} + \int_{0}^{t} (f(s) - \partial\phi_{\lambda}(w(s)))(t-s) ds.
$$

Then $u_\lambda$ satisfies

$T_{\lambda} u_{\lambda} = u_{\lambda}.$

Estimate $T_{\lambda}^{n}w - T_{\lambda}^{n}\hat{w}$ by a standard recurrence argument:

$$
|(T_{\lambda}w - T_{\lambda}(\hat{w})(t)| \le \frac{1}{\lambda} \frac{t^2}{2} \|w - \hat{w}\|_{C^o([0, t_0])} \quad \text{if } t \le t_0.
$$

Then

$$
|(T_{\lambda}^{n}w - T_{\lambda}^{n}\hat{w})(t)| \le \frac{t^{2n}}{(2n)! \lambda^n} \|w - \hat{w}\|_{C^{o}([0, t_0])} \quad \text{if } t \le t_0
$$

As we know that $T_{\lambda}^n w$ converges to $u_{\lambda}$ as $n \to \infty$, for any initial $w$, and on any compact time interval, we may write

$$
|u_{\lambda}(t) - w(t)| \le \sum_{n \ge 0} \| T_{\lambda}^{n+1} w - T_{\lambda}^{n} w \|_{C^{o}([0, t_0])} \le c h \frac{t}{\sqrt{\lambda}} \| T_{\lambda} w - w \|_{C^{o}([0, t_0])} \quad \text{if } t \le t_0.
$$

Take $w = v_\lambda$ defined by (15).

$$
(T_{\lambda} v_{\lambda} - v_{\lambda})(t) = u_{0} + t u_{1} - \int_{0}^{\lambda} \partial \phi_{\lambda}(v_{\lambda}(s))(t-s) \, ds - u_{0} - t u_{1} + \int_{0}^{t} (\partial \psi_{c})_{\lambda}(v_{\lambda}(s))(t-s) \, ds.
$$

If we assume $t_0 \le \rho/E$ and $\lambda \le \lambda_0$, we may apply the conclusions of Proposition 1 for $x = v_\lambda(t)$.
Therefore

$$
|(T_\lambda v_\lambda - v_\lambda)(t)| \le \int_0^t \left( k + \frac{E s \varepsilon(E s)}{\lambda} \right) (t-s) \, ds \le k \frac{t_0^2}{2!} + E \frac{t_0^3}{3! \lambda} \varepsilon(E t_0) \quad \text{if } t \le t_0,
$$

and finally we obtain the estimate (17).

The estimate (18) is a straightforward consequence of (17), thanks to the relation

$$
|\dot{u}_{\lambda}(t) - \dot{v}_{\lambda}(t)| \le \int_{0}^{\lambda} |(\partial \psi_{c})_{\lambda}(v_{\lambda}(t)) - \partial \phi_{\lambda}(v_{\lambda}(t))| dt + \int_{0}^{\lambda} |\partial \phi_{\lambda}(v_{\lambda}(t)) - \partial \phi_{\lambda}(u_{\lambda}(t))| dt.
$$

End of proof of Theorem 1, part b.

Denote $\tilde{u}_1 = 2P_c u_1 - u_1$. Clearly, $\tilde{u}_1$ is an element of $C$. Let us notice that, for any given $\eta > 0$, there exists a $\delta(\eta) > 0$ such that, if $|P_{K_0}v - u_0| \le \delta(\eta)$, then

$$
(v - P_{K_0}v, \tilde{u}_1) \le \eta |\tilde{u}_1| |v - P_{K_0}v|.
$$
If it were not the case, we could find a sequence $v_n$ with $Pv_n$ converging to $u_0$, and a strictly positive number $\eta_0$ such that

$$ (v_n - P_{\kappa_0} v_n, \tilde{u}_1) \geq \eta_0 |\tilde{u}_1| |v_n - P_{\kappa_0} v_n|. $$

We can see that $(v_n + Pv_n)/|v_n + Pv_n|$ converges to a certain $w$, which must be in $N$, and therefore, we obtain a contradiction. We may assume $\eta(\delta)$ to be an increasing function of $\delta$, such that

$$ |P_{\kappa_0}v - u_0| \le \delta \text{ implies } (v - P_{\kappa_0}v, \tilde{u}_1) \le \eta(\delta) |\tilde{u}_1| $$

Write now

$$ \begin{aligned} (u_{\lambda}(t), \tilde{u}_1) &= (u_{\lambda}(\pi\sqrt{\lambda}), \tilde{u}_1) + (\dot{u}_{\lambda}(\pi\sqrt{\lambda}), \tilde{u}_1)(t - \pi\sqrt{\lambda}) + \left( \int_{\pi\sqrt{\lambda}}^{t} (f(s) - A_{\lambda}u_{\lambda}(s))(t-s) \, ds, \tilde{u}_1 \right) \\ &\quad - \int_{\pi\sqrt{\lambda}}^{t} ([u_{\lambda}(s) - P_{\kappa_0}u_{\lambda}(s)], \tilde{u}_1)(t-s) \, ds \\ &\ge (u_{\lambda}(\pi\sqrt{\lambda}), \tilde{u}_1) + (\dot{u}_{\lambda}(\pi\sqrt{\lambda}), \tilde{u}_1)(t + \pi\sqrt{\lambda}) \\ &\quad + |f|_{L^2(\sqrt{\lambda})} \frac{t^{3/2}}{\sqrt{\lambda}} + k \frac{t^2}{2} - \eta(Et) \int_{\pi\sqrt{\lambda}}^{t} (1/\lambda) |u_{\lambda}(s) - P_{\kappa_0}u_{\lambda}(s)| |\tilde{u}_1| (t-s) \, ds \end{aligned} $$

From Proposition 1, and the estimate (11), we have:

$$ \int_0^T (1/\lambda) |u_\lambda(s) - P_{\kappa_0} u_\lambda(s)| \, ds \le C' + kT. $$

We obtain, in the limit as $\lambda \to 0$

$$ \begin{aligned} (u(t) - u_0, \tilde{u}_1) &\ge |\tilde{u}_1|^2 t - o(t), \text{ from where} \\ \left( \frac{d^+ u}{dt} (0), \tilde{u}_1 \right) &\ge |\tilde{u}_1|^2. \end{aligned} $$

On the other hand, as a result of (6), and of the fact that $\phi$ is continuous in a neighbourhood of $u_0$

$$ \left| \frac{d^+ u}{dt} (0) \right| = |u_1| = |\tilde{u}_1|. \text{ Conclusion (8) is now clear.} $$

This completes the proof of Theorem 1.

### 3. COUNTEREXAMPLES TO WELLPOSEDNESS

#### 3.a Discontinuity

The idea of this counterexample is quite simple.

Take $H = \mathbb{R}^2$, $K = \{(x_1, x_2) : x_1 \ge 0, x_1 + x_2 \ge 0\}$, $\phi = \psi_k$, $f = 0$; take as initial data $u_0^h = (1 - h, \frac{1}{2} + h)$, $u_1^h = (-1, -\frac{1}{2})$ where $|h| < \frac{1}{2}$. It is easy to check that the solutions are unique.

If $h = 0$, according to part (b) of Theorem 1, the solution is

$$ u^0(t) = (1 - t, \frac{1}{2}(1 - t)) \quad \text{if } t \le 1 $$

$$ u^0(t) = (\frac{1}{2}(t-1), t-1) \quad \text{if } t > 1. $$
If $h > 0$, the solution is

$$u^h(t) = (1 - h - t, \frac{1}{2} + h - \frac{1}{2}t) \quad \text{if } t \le 1$$

and it has a unique reflection at time $t = 1$ on the side $x_1 + x_2 = 0$ of K. Then

$$u^h(t) = (\frac{1}{2}(t-1) - h, t-1+h) \quad \text{if } t > 1.$$

If $h < 0$, the solution has a first reflection at time $t = 1 + 2h$; then $u^h(t) = (-3h - (t - 1 - 2h), (t - 1 - 2h)/2)$ until the time $1 + 4h$, when it has a second reflection, after which

$$u^h(t) = (-\frac{1}{2}t + \frac{1}{2} - h, t - 1 + h).$$

As $h$ tends to zero, remaining negative, $u^h$ tends to $\tilde{u}^0$ defined by

$$\tilde{u}^0(t) = (1 + t, \frac{1}{2}(1 - t)) \quad \text{if } t \le 1$$

$$\tilde{u}_0(t) = (-\frac{1}{2}(t-1), t-1) \quad \text{if } t > 1.$$

We can see now that, as a function of $h$, $u^h$ is right- but not left-continuous.

<figure><img src="image_3.png" /><figcaption>Fig. 2.</figcaption></figure>

### 3. b. Nonuniqueness

Two examples will be given, both of them with $\phi = \psi_K$.

$$\text{(i). Take } H = \mathbf{R}, K = \mathbf{R}^+, \phi = \psi_K, u_0 = u_1 = 0.$$

We seek an infinitely differentiable nonpositive $f$ which gives us two solutions. The first solution will be formed of an infinite number of arches, smaller and smaller as $t$ gets nearer to 0 (see figure below). The associated measure $\mu$ will then be atomic. The second solution will be $u = 0, \mu = f$.

More precisely, let $\rho \in \mathcal{D}(\mathbf{R})$ be an even, nonnegative function, the support of which is included in $[-1, +1]$, and satisfying $\int_{\mathbb{R}} \rho(x) dx = 1$. Denote $\rho_{\varepsilon}(x) = \rho(x/\varepsilon)$. Let $y_{\varepsilon} = 1_{[\varepsilon/2, 1 - \varepsilon/2]} * \rho_{\varepsilon/2}$ with $\varepsilon < \frac{1}{2}$. Clearly $y_{\varepsilon} = 1$ on $[\varepsilon, 1 - \varepsilon]$, $\text{supp } y_{\varepsilon} \subset [0, 1]$, and $y_{\varepsilon}$ is $C^{\infty}$. It is an easy exercise to check that

$$2 \int_{0}^{1} y_{\varepsilon}(s)(1-s) ds - \int_{0}^{1} y_{\varepsilon}(s) ds = 0.$$
<figure><img src="image_4.png" /><figcaption>Fig. 3.</figcaption></figure>

From now on we fix an $\epsilon$, and write $y_e = y$, $y_1 = \int_0^1 y(s) \, ds$ and $y_2 = \int_0^1 y(s)(1+s) \, ds$. Obviously $y_1 = 2y_2$. We look for an $f$ defined on each interval $[a_n, a_{n-1}]$ as follows: $f = 0$ on $[a_n, a_n + \sigma_n]$

$$f(t) = -\eta_n y\left(\frac{t - a_n - \sigma_n}{\tau_n - \sigma_n}\right) \text{ on } [a_n + \sigma_n, a_n + \tau_n] = [a_n + \sigma_n, a_{n-1}].$$

$f$ is $C^\infty$ except perhaps in 0, if it is defined.

We have necessarily

$$\ddot{u} = f \quad \text{on } [a_n, a_{n-1}]$$

$$u(a_n) = u(a_{n-1}) = 0$$

If we write $\frac{d^+ u}{dt}(a_n) = v_n$ then $\frac{d^- u}{dt}(a_{n-1}) = -v_{n-1}$.

Our aim is now to look for necessary conditions which will insure that $f$ has a meaning and is $C^\infty$, and $u$ has a meaning and is a solution of (P).

$$u(t) = v_n(t - a_n) \quad \text{on } [a_n, a_n + \sigma_n]$$

$$u(t) = v_n(t - a_n) - \eta_n \int_{a_n + \sigma_n}^{t} y(s)(t + s) \, ds \quad \text{on } [a_n + \sigma_n, a_{n-1}],$$

which gives us

$$v_n \tau_n = \eta_n y_2 (\tau_n - \sigma_n)^2$$

$$v_n + v_{n-1} = \eta_n y_1 (\tau_n - \sigma_n).$$

Let $v_{n-1} = \alpha_n v_n$. Then

$$\tau_n = \sigma_n \frac{\alpha_n + 1}{\alpha_n - 1} \quad \text{and} \quad \eta_n \sigma_n = \frac{v_n(\alpha_n^2 - 1)}{2y_1}.$$

Choose $\alpha_n = n$ for $n \ge 4$. This implies

$$\eta_n \sigma_n = \frac{v_4 (n^2 - 1) 4!}{2 y_1 n!} = \frac{12 v_4 n^2 - 1}{y_1 n!}.$$
Take

$$ \eta_n = 12 \frac{v_4}{y_1 (n-4)!} \quad \text{and} \quad \sigma_n = \frac{n+1}{n(n-2)(n-3)}. $$

For $u$ to be a solution of (P), we have only to check that

$$ \sum_{n \ge 4} \mu(\{a_n\}) = 2 \sum_{n \ge 4} v_n = 2 \sum_{n \ge 4} \frac{1}{n!} 24 v_4 < + \infty $$

$$ \sum_{n \ge 4} \tau_n = \sum_{n \ge 4} \frac{n+1}{n(n-2)(n-3)} \frac{n+1}{n-1} < + \infty. $$

Then $a_4 < + \infty$ and we can choose $T = a_4$.

For $f$ to be $C^\infty$ up to zero we need

$$ \eta_n / \left( \sum_{m \ge n-1} \tau_n \right)^r \sim \frac{1}{(n-4)!} (C_n)^p \frac{v_1}{2y_1} < + \infty \quad \forall p; $$

thus, we obtain the supplementary information that $f^{(p)}(0) = 0, \forall p \in \mathbb{N}$. Clearly $0$ with associated measure $\hat{\mu} = f$ is a solution. Thus we have built an infinitely differentiable $f$ such that (P) has two solutions.

(ii) Take $H = \mathbb{R}^2, f = 0$. We intend to build a convex set $K$ of $C^\infty$ boundary, such that with an initial velocity tangent to the boundary, we obtain two different solutions; one “along” the boundary, the other having an infinity of reflexions on it.

We shall use the results and notations of the preceding paragraph. Let us define a continuous parametrized curve with values in $C \approx \mathbb{R}^2$ as follows:

$$ F(t) = F(a_n) + e^{i\beta_n}(t - a_n) + e^{i(\beta_n - \pi/2)}u(t) \quad \text{if } t \in [a_n, a_{n-1}] \ 0 \le t \le T = a_4. $$

This curve will be continuously differentiable if

$$ e^{i\beta_n} \left( 1 - i \frac{d^+ u}{dt}(a_n) \right) = e^{i\beta_{n+1}} \left( 1 - i \frac{d^- u}{dt}(a_n) \right). $$

Then, necessarily $\beta_n - \beta_{n+1} = 2 \arctan v_n$. As $v_n = 24v_4/n!$, the series $\sum \beta_n$ converges. Moreover, if we choose $v_4$ small enough, $\sum_{n=4}^\infty \beta_n < \pi/2$, which means that $t \mapsto {}^nF(t)$ will have no tangent parallel to the imaginary axis. We have $F''(t) = e^{i(\beta_n - \pi/2)}f(t)$ on $[a_n, a_{n-1}]$. But $f$ is constructed in such a way that $f^{(p)}(a_n) = 0$ for all $p \ge 0, n \ge 4$. Therefore $F$ is $C^\infty$ on $[0, T]^p$. Moreover, the sign of the curvature of $F([0, T])$ is the sign of $\dot{F}_1 \ddot{F}_2 - \dot{F}_2 \ddot{F}_1$, where $F = F_1 + iF_2$.

We can easily check that $\dot{F}_1 \ddot{F}_2 - \dot{F}_2 \ddot{F}_1 = -f(t) \ge 0$. We can thus see that $F([0, T])$ is the boundary of a convex set $K$ of $C \approx \mathbb{R}^2$.

Take now initial conditions $u_0 = 0, \dot{u}_0 = a > 0$, and define a function $\psi_a$ on $[0, T]$ by

$$ \psi_a(t) = \frac{1}{a} \int_0^t |\dot{F}(\sigma)| \, d\sigma. $$

Let $\phi_a$ be the reciprocal of $\psi_a$.

Then $v(t) = F(\phi_a(t))$ is a solution of (P) on $[0, \psi_a(T)]$.

In fact, $v(t) \in K$, for all $t \in [0, \psi_a(T)]$.
$$ \frac{dv}{dt} = a \frac{\dot{F}(\phi_a(t))}{|\dot{F}(\phi_a(t))|} $$

$$ \frac{d^2 v}{dt^2} = a \frac{d}{dt} \left[ \frac{\dot{F}(\phi_a(t))}{|\dot{F}(\phi_a(t))|} \right]. $$

As $\dot{F}(\sigma) \neq 0$, for all $\sigma \in [0, T]$, $d^2u/dt^2$ is a bounded function which can be identified with a bounded measure. On the other hand, it is clear that $d^2v/dt^2$ is normal to $\partial K$, ($dv/dt$) is continuous at every point, $|dv/dt| = a$ and the initial conditions are satisfied. Therefore $u$ is a solution of (P) in the sense of Definition 1.

The other solution will be given by

$$ \tilde{v}(t) = F(a_n) + e^{i\beta_n}(at - a_n) \quad \text{on} \quad \left[ \frac{a_n}{a}, \frac{a_{n-1}}{a} \right] $$

Clearly, $\tilde{v}(t) \in K$ for all $t$. We have

$$ \frac{d^+ \tilde{v}}{dt} (a_n/a) = a e^{i\beta_n} $$

$$ \frac{d^- \tilde{v}}{dt} (a_n/a) = a e^{i\beta_{n+1}} $$

$$ \frac{d^2 \tilde{v}}{dt^2} = 0 \quad \text{if } t \in (a_n, a_{n-1}). $$

Therefore

$$ \frac{d^2 \tilde{v}}{dt^2} = a \sum_{n \ge 1} \delta_{a_n/a} (e^{i\beta_n} - e^{i\beta_{n+1}}) = -\tilde{\mu}. $$

It is easy to check, using the interpretation of the measure associated with a solution of (P), (13), (14), that $\tilde{v}$ is a solution of (P); clearly, the energy is conserved, and the initial conditions are satisfied.

*Comments on this part.* These results are closely connected with results on the propagation of singularities of a hyperbolic problem on a manifold with boundary. The first one obviously refers to diffraction in an angle. The second one was figured out by thinking of a tennis player who succeeds in making his ball bounce higher and higher, from a rest position, only by hitting it downwards. Professor L. Amerio told me that he had built an analogous example. The third type of counterexample is given, with a different construction by M. E. Taylor in his paper [4] concerning reflection of singularities of wave equations in an exterior domain of $\mathbb{R}^N$. When the complement of this domain is not convex, Taylor's theorem fails as there appear phenomena of nonuniqueness of the wave front set. I cannot see, presently, how all these results could be taken in account together in reasonable mathematical theory.

### 4. A UNIQUENESS THEOREM IN A PARTICULAR CASE

**THEOREM 2.** Let $\phi = \psi_K$, $f = 0$. Assume that the boundary $\partial K$ of the closed convex set $K$ is of class $C^3$, and that its gaussian curvature is strictly positive. Then the problem (P) admits a unique solution on $[0, +\infty]$ in the sense of Definition 1. Moreover, if the initial data $u_0$ is given
on $\partial K$ and $u_1$ is tangent to $\partial K$, then $u$ runs along the geodesic of $\partial K$ passing through $u_0$ and tangent to $u_1$ with the speed $|u_1|$. If the initial data are not such, then $u$ is never tangent to $\partial K$, and it has a finite number of reflections in a finite time.

*Proof.* Denote by $\mathcal{S}_K(u)$ the tangent cone to $K$ at $u$, i.e. $\mathcal{T}_K(u) = \bigcup_{\tau>0} \tau(K - u)$ and by $n(u)$ the exterior unit normal to $K$ at $u$. Remark that

$$ \mathcal{S}_K(u) = \{v: (v, n(u)) \ge 0\} \quad \text{for all } u \in \partial K. $$

Let us first prove the local uniqueness at every point $t$. If $u(t) \in \text{int } K$, or if $u(t) \in \partial K$ and $(d^-u/dt)(t)$ is the interior of $\mathcal{S}_K(u(t))$, the local uniqueness is immediate.
If $u(t) \in \partial K$, and $d^-u/dt(t) \notin \mathcal{S}_K(u(t))$, then

$$ \left| \frac{d^+ u}{dt}(t) \right| = \left| \frac{d^- u}{dt}(t) \right|. $$

and

$$ \frac{d^+ u}{dt}(t) - \frac{d^- u}{dt}(t) = kn(u(t)) $$

according to Theorem 1.
Necessarily the vector

$$ \frac{d^+ u}{dt}(t) = \frac{d^- u}{dt}(t) - 2\left(\frac{d^- u}{dt}(t), n(u(t))\right) n(u(t)) $$

is in the interior of $\mathcal{S}_K(u(t))$. The local uniqueness is then clear. The only difficult case is when $u(t) \in \partial K$, and $((d^-u/dt)(t), n(u(t))) = 0$. We need the following result.

**LEMMA 2.** If $\partial K$ is of class $C^3$ and its gaussian curvature is strictly positive, then there exists no sequence of points of $\partial K$, $(u_m)_{m \in \mathbb{N}}$ converging to $u_\infty$, such that $u_m$ is a reflexion point for all $m$, i.e.

$$ \left\{ \begin{array}{l} \frac{(u_m - u_{m-1}, n(u_m))}{|u_m - u_{m-1}|} = \frac{(u_m - u_{m+1}, n(u_m))}{|u_m - u_{m+1}|} \quad \text{and} \\ \\ \frac{u_m - u_{m-1}}{|u_m - u_{m-1}|} - \frac{(u_m - u_{m-1}, n(u_m)) n(u_m)}{|u_m - u_{m-1}|} = \frac{u_{m+1} - u_m}{|u_{m+1} - u_m|} - \frac{(u_{m+1} - u_m, n(u_m)) n(u_m)}{|u_{m+1} - u_m|} \end{array} \right. $$

such that $u_{m+1} - u_m$ has a limit direction, and such that

$$ \sum_{m \in \mathbb{N}} |u_{m+1} - u_m| < + \infty. $$

Let us show that this lemma implies local uniqueness. With $u_0 \in \partial K$, $(n(u(t_0)), (d^-u/dt)(t_0)) = 0$, then necessarily $((d^+u/dt)(t_0), n(u(t_0))) = 0$. Suppose that there is no right neighbourhood of $t_0$ such that $u(t)$ belongs to $\partial K$ in all this neighbourhood. Then we can find a $t_1$, arbitrarily near $t_0$, such that $u(t_1) \notin \partial K$. There must exist at most a finite number of reflections between $t_0$ and $t_1$, in order to satisfy Lemma 2. Then $((d^+u/dt(t_0)), n(u(t_0))) < 0$ and we obtain a contradiction. Therefore there exists an $\eta > 0$ such that $u(t) \in \partial K$ for $t_0 \le t \le t_0 + \eta$.
Let $\mu$ be the measure associated with $u$, and define a real valued measure $(\mu, z)$ with $z \in C^0([0, T]; H)$ by

$$ \langle (\mu, z), \phi \rangle = \langle \mu, z \phi \rangle \quad \forall \phi \in C^0([0, T]). $$

Let $s = (w - (w, n(u))n(u)) \chi$ with $w \in H$, and $\chi \in C^0([0, T])$, supp $\chi \subset [t_0, t_0 + \eta]$. It is clear that $s(t) \in \mathcal{S}_K(u(t))$, and $s$ is universally integrable on $[0, T]$. According to the interpretation of $\mu$, (13) and (14), necessarily $\langle \mu, s \rangle = 0$.

So

$$ (\mu, w)|_{(t_0, t_0 + \eta)} = (n(u), w)(\mu|_{(t_0, t_0 + \eta)}, n(u)). $$

If we set $v = (\mu|_{(t_0, t_0 + \eta)}, n(u))$, we may then identify $\mu|_{(t_0, t_0 + \eta)}$ and $n(u)v$. Suppose $u(t_0) = 0$, which does not restrict the generality, and denote $n(t_0) = n_0$, $u_N = -(u, n_0)$, $u' = u - (u, n_0)n_0$. Represent $\partial K$ in a neighbourhood of 0 as follows:

$$ |u| \le \alpha \quad \text{and} \quad u_N = f(u') \quad \text{imply} \quad u \in \partial K. $$

Here $f$ is convex, twice continuously differentiable, and $Df(0) = 0$. We differentiate the relation $u_N(t) = f(u'(t))$ on $[t_0, t_0 + \eta]$.

$$ (19) \qquad \frac{du_N}{dt} = Df(u'(t)) \frac{du'}{dt}; $$

$\frac{du}{dt}$ is continuous, as $\left|\frac{d^+ u}{dt}\right| = \left|\frac{d^- u}{dt}\right|$, $\frac{d^+ u}{dt} - \frac{d^- u}{dt} = kn(u)$ (for a certain real $k$), and

$$ \left( \frac{d^+ u}{dt}, n(u) \right) = \left( \frac{d^- u}{dt}, n(u) \right) = 0. $$

We now differentiate (19) in the sense of distributions:

$$ (20) \qquad \frac{d^2 u_N}{dt^2} = D^2 f(u'(t)) \frac{du'}{dt} \frac{du'}{dt} + Df(u'(t)) \frac{d^2 u'}{dt^2}. $$

Explicitly

$$ n(u) = \frac{(Df(u'); -1)}{(1 + |Df(u')|^2)^{1/2}}, $$

whence

$$ \frac{d^2 u'}{dt^2} = \frac{-v Df(u')}{(1 + |Df(u')|^2)^{1/2}}, $$

$$ \frac{d^2 u_N}{dt^2} = \frac{v}{(1 + |Df(u')|^2)^{1/2}}, $$

$$ \left( Df(u'), \frac{d^2 u'}{dt^2} \right) = - \frac{v |Df(u')|^2}{(1 + |Df(u')|^2)^{1/2}}. $$

Substituting in relation (20) we obtain

$$ v(1 + |Df(u')|^2)^{1/2} = D^2 f(u') \frac{du'}{dt} \frac{du'}{dt}. $$
Therefore, $v$ can be identified with a function and

$$ \frac{d^2 u}{dt^2} + \frac{D^2 f(u') \frac{du'}{dt} \frac{du'}{dt}}{(1 + |Df(u')|^2)^{1/2}} n(u) = 0. $$

This is precisely the equation of the geodesics of $\partial K$. As $\partial K$ is of class $C^3$, we know there is uniqueness. By a classical argument, local uniqueness implies global uniqueness.

To complete the proof of Theorem 2, we need only to establish Lemma 2.

*Proof of Lemma 2.* Suppose there exists a sequence $u_m \in \partial K$ such that

(i) $u_m \to u_\infty$ as $m \to \infty$,

(ii) $$ \frac{Q_m(u_m - u_{m+1})}{|u_m - u_{m+1}|} = \frac{Q_m(u_m - u_{m-1})}{|u_{m-1} - u_m|}, $$

(iii) $$ \frac{P_m(u_m - u_{m+1})}{|u_m - u_{m+1}|} = \frac{P_m(u_{m-1} - u_m)}{|u_{m-1} - u_m|}. $$

where $Q_m x = -(x, n(u_m))$, $P_m x = x + n(u_m) Q_m x$.

In a neighbourhood $U$ of $u_\infty$ we have the representations of $\partial K$ given by

$$ u \in \partial K \cap U \Rightarrow Q_m(u - u_m) = f_m(P_m(u - u_m)), $$

where $f_m$ is of class $C^3$ and $Df_m(0) = 0$; $f_m$ is convex.

Obviously

(21) $$ \frac{f_m(P_m(u_{m-1} - u_m))}{|P_m(u_{m-1} - u_m)|} = \frac{f_m(P_m(u_{m+1} - u_m))}{|P_m(u_{m+1} - u_m)|}. $$

On the other hand,

(22) $$ f_m(u') = D^2 f_m(0) u'u' + |u'|^2 g_m(u') \quad \text{if } (u', n(u_m)) = 0, $$

with

$$ |g_m(u')| \le h(|u'|). $$

The function $h$ does not depend on $m$; $\lim_{r \to 0} h(r) = 0$, and $h$ is increasing

$$ \begin{aligned} \text{Set } \kappa_m &= D^2 f_m(0) \left( \frac{P_m(u_{m-1} - u_m)}{|P_m(u_{m-1} - u_m)|} \right) \left( \frac{P_m(u_{m-1} - u_m)}{|P_m(u_{m-1} - u_m)|} \right) \\ &= D^2 f_m(0) \left( \frac{P_m(u_{m+1} - u_m)}{|P_m(u_{m+1} - u_m)|} \right) \left( \frac{P_m(u_{m+1} - u_m)}{|P_m(u_{m+1} - u_m)|} \right). \end{aligned} $$

From (21) and (22) we obtain:

(23) $$ \kappa_m \left( |P_m(u_{m-1} - u_m)| - |P_m(u_{m+1} - u_m)| \right) + |P_m(u_{m-1} - u_m)| g_m \left( P_m(u_{m-1} - u_m) \right) \\ - |P_m(u_{m+1} - u_m)| g_m \left( P_m(u_{m+1} - u_m) \right) = 0 $$

Since we have assumed that the gaussian curvature of $\partial K$ at $u_\infty$ is strictly positive, $\kappa_m \ge \kappa_0 > 0$ for $m$ large enough. Dividing both sides of (23) by $|P_m(u_{m-1} - u_m)|$ we obtain
$$
\kappa_0 \left| \frac{P_m(u_{m+1} - u_m)}{P_m(u_{m-1} - u_m)} \right| - 1 \le \left| \frac{P_m(u_{m+1} - u_m)}{P_m(u_{m-1} - u_m)} \right| - 1 \left| h\left(P_m(u_{m+1} - u_m)\right) + h\left(P_m(u_{m-1} - u_m)\right) \right| + h\left(P_m(u_{m+1} - u_m)\right).
$$

If we choose $m_0$ such that

$$
m \ge m_0 \Rightarrow h(|P_m(u_{m+1} - u_m)|) \le \kappa_0/2,
$$

then

$$
\frac{\kappa_0}{2} \left| \frac{P_m(u_{m+1} - u_m)}{P_m(u_{m-1} - u_m)} \right| - 1 \le h\left(\left|P_m(u_{m-1} - u_m)\right|\right) + h\left(\left|P_m(u_{m+1} - u_m)\right|\right).
$$

Set $l_m = |u_{m-1} - u_m|$. Finally

$$
(24) \quad \frac{\kappa_0}{2} \left| \frac{P_m(u_{m+1} - u_m)}{P_m(u_{m-1} - u_m)} \right| - 1 \le h(l_m) + h(l_{m+1}).
$$

On the other hand we have

$$
P_m(u_{m-1} - u_m) - f_m(P_m(u_{m-1} - u_m)) n(u_m) = u_{m-1} - u_m.
$$

Therefore

$$
|l_m - P_m(u_{m-1} - u_m)| \le |D^2 f_m(0)(P_m(u_{m-1} - u_m))(P_m(u_{m-1} + u_m))| + \\ \qquad + |P_m(u_{m-1} - u_m)|^2 h(|P_m(u_{m-1} - u_m)|).
$$

So we have a constant K such that

$$
(25) \qquad \left| \frac{l_m}{P_m(u_{m-1} - u_m)} - 1 \right| \le K l_m
$$

and analogously

$$
(26) \qquad \left| \frac{l_{m+1}}{P_m(u_{m+1} - u_m)} - 1 \right| \le K l_{m+1}.
$$

From (24), (25) and (26) we obtain

$$
| l_{m+1} / l_m - 1 | \le K'(h(l_m) + h(l_{m+1}) + l_m + l_{m+1}).
$$

Set $\tilde{h}(r) = K'(h(r) + r)$, then

$$
l_{m+1} \ge l_m + l_m(\tilde{h}(l_{m+1}) + \tilde{h}(l_m)).
$$

Since $l_m$ converges to 0, we can find an arbitrarily large $m_0$ such that $l_m \le l_{m_0}$, for all $m \ge m_0$.

Then

$$
l_{m+1} \ge l_m - 2l_m \tilde{h}(l_{m0}) \quad \text{if } m \ge m_0.
$$

$$
\sum_{m \ge m_0+1} l_m \ge \sum_{m \ge m_0} l_m (1 - 2\tilde{h}(l_{m0})) = \sum_{m \ge m_0+1} (1 - 2\tilde{h}(l_{m0})) l_m + (1 - 2\tilde{h}(l_{m0})) l_{m0}.
$$

Therefore

$$
\sum_{m \ge m_0 + 1} l_m \ge \frac{1 - 2\tilde{h}(l_{m0})}{2\tilde{h}(l_{m0})} \cdot l_{m0}.
$$
It is clear that we can choose $m_0$ so large that the right hand side of this inequality is arbitrarily large. We obtain therefore a contradiction, as $u_{m+1} - u_m$ is supposed to have a limit direction when $m$ tends to infinity. This completes the proof of Theorem 2.

The main results of this paper have been stated in ref. [6].

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2. ROCKAFELLAR R. T., *Convex Analysis*. Princeton Univ. Press (1970).

3. ZARANTONELLO E. H., *Contributions to Nonlinear Functional Analysis*. pp. 237–424. Academic Press (1971).

4. TAYLOR M. E., *Communs pure appl. Math. 38*, 1–38 (1976).

5. ROCKAFELLAR R. T., *Pacif. J. Math. 39*, 439–469 (1971).

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