# A boundary thin obstacle problem for a wave equation

Jong Uhn Kim

► To cite this version:

Jong Uhn Kim. A boundary thin obstacle problem for a wave equation. Communications in Partial Differential Equations, 1989, 14 (8-9), pp.1011-1026. 10.1080/03605308908820640. hal-01330421

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Distributed under a Creative Commons Attribution 4.0 International License
A BOUNDARY THIN OBSTACLE PROBLEM FOR A WAVE EQUATION

BY JONG UHN KIM

Department of Mathematics

Virginia Polytechnic Institute & State University

Blacksburg, VA 24061

0. Introduction.

In this paper we shall discuss a hyperbolic variational inequality associated with the following
initial-boundary value problem:

$$
\frac{\partial^2 u}{\partial t^2} - \Delta u = f \text{ in } \Omega \times (0, T), \quad (0-1)
$$

$$
u(x, 0) = u_0(x), \quad \partial_t u(x, 0) = u_1(x) \text{ in } \Omega, \qquad (0-2)
$$

$$
u = 0 \text{ on } \Gamma_1 \times (0, T), \qquad (0-3)
$$

$$
u(x,t) \ge \phi(x) \text{ on } \Gamma_2 \times (0,T), \qquad (0-4)
$$

$$
\frac{\partial u}{\partial \nu} \ge 0, \quad (u - \phi) \frac{\partial u}{\partial \nu} = 0 \text{ on } \Gamma_2 \times (0, T). \qquad (0-5)
$$

Here Ω is a bounded open subset of Rⁿ, n ≥ 2, with smooth boundary ∂Ω ∈ C². We assume
that Γ₁ and Γ₂ are disjoint, nonempty open subsets of ∂Ω such that ∂Ω = Γ₁ ∪ Γ₂ = Γ₁ ∪ Γ₂ and
∂Γ₁ = ∂Γ₂ ∈ C². The above functions f(x, t), u₀(x), u₁(x) and φ(x) are given, and ∂²φ/∂ν² denotes the
outward normal derivative on ∂Ω. We assume that φ ≤ 0 and u₀ ≥ φ on Γ₂. We can put the above
problem in the form of a variational inequality: find u(x, t) which satisfies (0-2), (0-3), (0-4) and

$$
\langle \frac{\partial^2 u}{\partial t^2}, w - u \rangle + a(u, w - u) \geq \langle f, w - u \rangle \quad (0-6)
$$
for every $w \in G$, for all $t \in (0, T)$. Here we use the notation:

$$ 
\begin{aligned}
<u, w> &= \int_{\Omega} u w dx, \\
a(u, w) &= \int_{\Omega} \nabla u \cdot \nabla w dx, \\
G &= \{w \in H^1(\Omega) : w = 0 \text{ on } \Gamma_1 \text{ and } w \ge \phi \text{ on } \Gamma_2\}.
\end{aligned}
 $$

If a solution $u$ of this new problem is sufficiently smooth, one can show that $u$ also satisfies (0-1) and (0-5) so that $u$ is a solution of the original problem (0-1) through (0-5). The corresponding stationary variational inequality: find $u \in G$ which satisfies

$$ a(u, w - u) \ge <f, w - u>, \text{ for all } w \in G, \qquad (0-7) $$

is called the boundary thin obstacle problem (or the Signorini problem), and its mathematical theory has been well-established: see [6] and references therein. However, to the author's knowledge, nothing has been known about the solution of (0-1) through (0-5). The purpose of this paper is to establish the existence of a (weak) solution of (0-1) through (0-5). This problem is an example of a hyperbolic variational inequality where a constraint is imposed on the unknown function rather than on the time derivative of the unknown function. Hyperbolic variational inequalities have been extensively investigated in the case where a constraint is imposed on the time derivative of the unknown function; see [3] and [11] among others. On the other hand, when a constraint is imposed on the unknown function, existence theorems seem to be known mainly for special problems of space dimension one; see [1], [2], [5], [8], [12], [13] and [14]. In the case of several space dimensions, [10] and this paper seem to be the only known works. [10] resolved the above problem by microlocal techniques when $\Omega$ is a half-space; but the same method does not extend to a general domain, as is explained in [10].

In this paper, the basic procedure to prove the existence of a solution consists of a penalty method [11], a multiplier technique [9] and compensated compactness [4]. Finally, we note that the questions of uniqueness and energy conservation for solutions of our problem are still open.

1. Notation and Preliminaries.

$$ \partial_t = \frac{\partial}{\partial t}, \partial_{tt} = \frac{\partial^2}{\partial t^2}, \nabla g = \left( \frac{\partial g}{\partial x_1}, \dots, \frac{\partial g}{\partial x_n} \right). $$

When $E$ is an open subset of $R^N$, $N \ge 1$, $D'(E)$ denotes the space of distributions over $E$ and $C_0^m(E) = \{w \in C^m(E) : \text{supp } w \text{ is a compact subset of } E\}$. We assume that $\partial\Omega \in C^2$ and $\partial\Gamma_1 = \partial\Gamma_2 \in C^2$. For $m$=nonnegative integer,
$$
H^m(\Omega) = \{w \in L^2(\Omega) : \left(\frac{\partial}{\partial x_1}\right)^{\alpha_1} \cdots \left(\frac{\partial}{\partial x_n}\right)^{\alpha_n} w \in L^2(\Omega), 0 \le \alpha_1 + \cdots + \alpha_n \le m\},
$$

$$
H_0^m(\Omega) = \text{the completion of } C_0^\infty(\Omega) \text{ in } H^m(\Omega),
$$

$$
H^{-m}(\Omega) = \text{the dual of } H_0^m(\Omega),
$$

$$
H_{\Gamma_1}^{1}(\Omega) = \{w \in H^1(\Omega) : w = 0 \text{ on } \Gamma_1\}.
$$

For $s$=nonnegative real, $H^*(\Omega)$ is defined by interpolation and $H^{-s}(\Omega)$=the dual of $H_0^*(\Omega)$. We denote by $<,>$ the duality pairing between $H^{-\alpha}(\Omega)$ and $H^{\alpha}(\Omega)$, $0 \le \alpha \le \frac{1}{2}$. Then, for $g \in L^2(\Omega)$, $h \in H^\alpha(\Omega)$, $0 \le \alpha \le \frac{1}{2}$, we have $<g, h> = \int_\Omega g h dx$.

As above, the bilinear form $a(\cdot, \cdot)$ is defined by $a(g, h) = \int_{\Omega} \nabla g \cdot \nabla h \, dx$ and

$$
G = \{ w \in H_1^1(\Omega) : w \ge \phi \text{ on } \Gamma_2 \},
$$

where $\phi(x) \in C^1(\bar{\Omega})$ and $\phi(x) \le 0$ on $\Gamma_2$.

Next we define the distance function $d$ by $d(x)=\text{dist}(x,\partial\Omega)$ and set

$$
V_{\delta} = \{ x \in \bar{\Omega} : d(x) < \delta \}.
$$

Let $\delta_0$ be a positive number such that $1/\delta_0$ bounds the principal curvatures of $\partial\Omega$. Since $\Omega$ is bounded and $\partial\Omega \in C^2$, it is known that $d(x) \in C^2(V_{\delta_0})$; see [7]. We also set

$$
U_{\delta} = \{ x \in \Omega : d(x) < \delta \},
$$

$$
S_{\delta} = \{ x \in \Omega : d(x) = \delta \}.
$$

Then, $S_\delta$ is a $C^2$-surface for $0 \le \delta < \delta_0$. We shall use the following lemmas later.

LEMMA 1.1. There is a function $h(x) \in C^1(\bar{\Omega})^n$ such that $h(x)=\text{the outward unit normal vector at $x$ on $S_\delta$ if } d(x) = \delta \le \delta_0/2 \text{ and } h(x) = 0 \text{ if } d(x) \ge \delta_0$.

PROOF: For each $x \in S_\delta$, $0 \le \delta < \delta_0$, there is a unique point $y(x) \in \partial\Omega$ such that $\text{dist}(x, y(x)) = \delta$
and the outward normal vector at $y(x)$ on $\partial\Omega$ is also outward normal at $x$ on $S_\delta$. Furthermore,
$y(x) \in C^1(V_{\delta_0})$. Next we denote by $\nu(y)$ the outward unit normal vector at $y$ on $\partial\Omega$. Then, $\nu(y)$
is a $C^1$ function of $y$. Let us choose a function $\rho(t) \in C^1(\mathbb{R})$ such that $\rho(t) = 1$ for $t \le \delta_0/2$ and
$\rho(t) = 0$ for $t > \frac{3}{2}\delta_0$. We then set $h(x) = \nu(y(x))\rho(d(x))$, which achieves our goal.

The next lemma is known and stated without proof.

LEMMA 1.2. Let $g \in L^1(\Omega)$. Then, for $0 < \delta^* < \delta_0$,

$$
\int_{U_{\delta}^{*}} g \, dx = \int_{0}^{\delta^{*}} \int_{S_{\delta}} g \, d\sigma_{\delta} \, d\delta \quad (1-1)
$$

holds where $d\sigma_\delta$ is the surface element on $S_\delta$.
Next we write a well-known consequence of the Div-Curl lemma of compensated compactness.

LEMMA 1.3. Let {$u_k$} be a sequence of functions such that

$$ \|u_k\|_{L^2(0,T;H^1(\Omega))} \le M, \quad (1-2) $$

$$ \|\partial_t u_k\|_{L^2(0,T;L^2(\Omega))} \le M, \quad (1-3) $$

where $M$ is a positive constant independent of $k$ and each $u_k$ satisfies

$$ \frac{\partial^2 u_k}{\partial t^2} - \Delta u_k = g_k \quad (1-4) $$

where {$g_k$} is a bounded sequence in $L^2(\Omega \times (0,T))$, and, as $k \to \infty$,

$$ u_k \to u \text{ weakly in } L^2(0,T; H^1(\Omega)). \quad (1-5) $$

Then, as $k \to \infty$

$$ \left(\frac{\partial u_k}{\partial t}\right)^2 - |\nabla u_k|^2 \to \left(\frac{\partial u}{\partial t}\right)^2 - |\nabla u|^2 \quad (1-6) $$

in $D'(\Omega \times (0,T))$.

PROOF: Set $v_{k_i} = \frac{\partial u}{\partial x_i}$, $w_{k_i} = \frac{\partial u}{\partial x_i}$, $i = 1, \dots, n$, and $v_{k(n+1)} = -\frac{\partial u}{\partial t}$, $w_{k(n+1)} = \frac{\partial u}{\partial t}$. Then, apply Corollary 4.3 of [4].

Finally, we state

LEMMA 1.4. Let {$u_k$} be a sequence of functions such that as $k \to \infty$,

$$ u_k \to u \text{ weakly in } L^\infty(0,T; H^\beta(\Omega)) \quad (1-7) $$

$$ \partial_t u_k \to \partial_t u \text{ weakly in } L^2(0,T; H^\alpha(\Omega)) \quad (1-8) $$

where $-1 \le \alpha < \beta \le 1$. Then, we have

$$ u_k \to u \text{ in } C([0,T]; H^r(\Omega)) \quad (1-9) $$

for any $r < \beta$.

PROOF: A special case was proved in [8]. The same argument can still be employed. Without loss of generality, we may assume $\alpha \ne -\frac{1}{2}$, $\beta \ne -\frac{1}{2}$, because we can take smaller $\alpha$ or $\beta$ if necessary.

Let $r = \theta\alpha + (1-\theta)\beta$, $0 < \theta < 1$, $r \ne -\frac{1}{2}$, $\frac{1}{2}$. There is $E \subset [0,T]$ such that $\mathrm{meas}(E) = 0$ and for all $t_1, t_2 \in [0,T] \setminus E$, $t_1 < t_2$,
$$
\begin{align*}
& \|u_k(t_2) - u_k(t_1)\|_{H^r(\Omega)} \le \\
& \le M_{\theta} \|u_k(t_2) - u_k(t_1)\|_{H^\alpha(\Omega)}^{\theta} \|u_k(t_2) - u_k(t_1)\|_{H^{\beta}(\Omega)}^{1-\theta}, \\
& \le M_{\theta} \left( \int_{t_1}^{t_2} \|\partial_t u_k(t)\|_{H^\alpha(\Omega)} dt \right)^{\theta} \le M_{\theta} (t_2 - t_1)^{\theta/2}, \tag{1-10}
\end{align*}
$$

holds for all $k$ and $0 < \theta < 1$ such that $\theta\alpha + (1-\theta)\beta \neq \frac{1}{2}, -\frac{1}{2}$, where $M_\theta$ denotes positive constants independent of $t_1, t_2$ and $k$. Consequently, each $u_k$ belongs to $C([0,T]; H^r(\Omega))$ for every $\alpha < r < \beta$ possibly after a modification on $E$. Since the embedding $H^{r2}(\Omega) \subset H^{r1}(\Omega)$ is compact for every $r_1 < r_2$, we can use the Ascoli theorem to arrive at the conclusion.

## 2. Approximate Problem.

In this section we shall consider an approximate problem with a penalty parameter $k > 0$. The problem is: find a function $u_k(x, t)$ such that

$$ \partial_{tt} u_k(x, t) \in L^\infty(0, T; L^2(\Omega)), \partial_t u_k \in L^\infty(0, T; H_1^r(\Omega)) \quad (2-1) $$

$$ u_k(x, 0) = u_{0k}(x), \partial_t u_k(x, 0) = u_{1k}(x) \quad (2-2) $$

$$ < \partial_{tt} u_k, w > + a(u_k, w) = < f_k, w > + \int_{\Gamma_2} (k(u_k - \phi)^{-} - \frac{1}{k} \partial_t u_k) w d\sigma \quad (2-3) $$

holds for all $w \in H_1^r(\Omega)$, for almost all $t \in (0, T)$. Here $d\sigma$ is the surface element on $\partial\Omega$ and $g^- = \max(0, -g)$.

Our strategy consists of three steps.

(Step 1) Establish the existence of solution of (2-1), (2-2) and (2-3).

(Step 2) Obtain a priori estimates independent of $k > 0$.

(Step 3) Pass $k \to \infty$ to arrive at a solution of the original problem.

In this section, the first two steps will be carried out.

### 2.1 Existence of solution

Throughout this subsection, we shall suppress the subscript $k$.

**PROPOSITION 2.1.** Let $u_0, u_1 \in C_0^\infty(\Omega)$, $f \in C^1([0, T]; L^2(\Omega))$ and $\phi \in C^1(\bar{\Omega})$ with $\phi \le 0$ on $\Gamma_2$. Then there is a solution of (2-1), (2-2) and (2-3).

**PROOF:** We set $u = v + u_0 + tu_1$. Then,

$$ v(x, 0) \equiv 0, \partial_t v(x, 0) \equiv 0, \quad (2-4) $$

and (2-3) is equivalent to

$$
\begin{align}
& < \partial_{tt} v, w > + a(v, w) = < f, w > \nonumber \\
& + < \Delta(u_0 + tu_1), w > + k \int_{\Gamma_2} (v - \phi)^{-} w d\sigma \nonumber \\
& - \frac{1}{k} \int_{\Gamma_2} (\partial_t v) w d\sigma, \tag{2-5}
\end{align}
$$
for all $w \in H_1^1(\Omega)$, for almost all $t \in (0, T)$. We shall employ the Galerkin approximation procedure. Let $\{w_j\}_{j=1}^\infty$ be a basis for $H_1^1(\Omega)$ where each $w_j \in C^\infty(\bar{\Omega}) \cap H_1^1(\Omega)$. Let us set

$$v_m = \sum_{j=1}^{m} a_{mj}(t)w_j(x) \qquad (2-6)$$

where each $a_{mj}(t)$ is determined from

$$a_{mj}(0) = \dot{a}_{mj}(0) = 0, \qquad (2-7)$$

and

$$ \begin{aligned} & \langle \partial_{tt} v_m, w_j \rangle + a(v_m, w_j) = \langle f, w_j \rangle \\ & + \langle \Delta(u_0 + tu_1), w_j \rangle + k \int_{\Gamma_2} (v_m - \phi)^{-} w_j d\sigma \\ & - \frac{1}{k} \int_{\Gamma_2} (\partial_t v_m) w_j d\sigma, \quad j = 1, \dots, m. \end{aligned} \qquad (2-8) $$

For each $m$, there is a unique set of functions $a_{mj}(t) \in C^2([0, T])$, $j = 1, \dots, m$, which satisfy (2-7) and (2-8). We next obtain estimates independent of $m$. Let us substitute $\partial_t v_m$ for $w_j$ in (2-8):

$$ \begin{aligned} & \frac{1}{2} \frac{d}{dt} \|\partial_t v_m\|_{L^2(\Omega)}^2 + \frac{1}{2} \frac{d}{dt} a(v_m, v_m) \\ & + \frac{1}{2} k \frac{d}{dt} \|(v_m - \phi)^{-}\|_{L^2(\Gamma_2)}^2 \\ & = \langle f, \partial_t v_m \rangle + \langle \Delta(u_0 + tu_1), \partial_t v_m \rangle - \frac{1}{k} \|\partial_t v_m\|_{L^2(\Gamma_2)}^2. \end{aligned} \qquad (2-9) $$

Note that $v_m - \phi \in C^1([0, T] \times \bar{\Omega})$ and hence, $(v_m - \phi)^{-}$ is Lipschitz continuous and

$$\frac{d}{dt} \|(v_m - \phi)^{-}\|_{L^2(\Gamma_2)}^2 \in L^{\infty}(0, T).$$

It follows from (2-7) and (2-9) that

$$\|\partial_t v_m\|_{L^2(\Omega)} \le M, \text{ for all } t \in [0, T] \qquad (2-10)$$

$$\|v_m\|_{H_1^1(\Omega)} \le M, \text{ for all } t \in [0, T] \qquad (2-11)$$

$$\|(v_m - \phi)^{-}\|_{L^2(\Gamma_2)} \le M, \text{ for all } t \in [0, T] \qquad (2-12)$$

$$\int_{0}^{T} \| \partial_t v_m \|_{L^2(\Gamma_2)}^2 dt \le M, \qquad (2-13)$$

where $M$ denotes positive constants independent of $m$. Next we differentiate (2-8) with respect to $t$ and substitute $\partial_{tt} v_m$ for $w_j$:
$$
\begin{align*}
& < \partial_{ttt} v_m, \delta_{tt} v_m > + a(\partial_t v_m, \partial_{tt} v_m) \\
& = < \partial_t f, \partial_{tt} v_m > + < \Delta u_1, \partial_{tt} v_m > \\
& \quad + k \int_{\Gamma_2} \partial_t \{(v_m - \phi)^-\} \partial_{tt} v_m d\sigma \\
& \quad - \frac{1}{k} \int_{\Gamma_2} (\partial_{tt} v_m)^2 d\sigma, \tag{2-14}
\end{align*}
$$

which yields

$$
\begin{align*}
& \frac{1}{2} \frac{d}{dt} \| \partial_{tt} v_m \|_{L^2(\Omega)}^2 + \frac{1}{2} \frac{d}{dt} a(\partial_t v_m, \partial_t v_m) \\
& \le \| \partial_t f \|_{L^2(\Omega)} \| \partial_{tt} v_m \|_{L^2(\Omega)} + \| \Delta u_1 \|_{L^2(\Omega)} \| \partial_{tt} v_m \|_{L^2(\Omega)} \\
& \quad + \frac{1}{2} k^3 \| \partial_t \{(v_m - \phi)^-\} \|_{L^2(\Gamma_2)}^2 - \frac{1}{2k} \| \partial_{tt} v_m \|_{L^2(\Gamma_2)}^2. \tag{2-15}
\end{align*}
$$

We consider (2-8) at $t=0$ and substitute $\partial_{tt}v_m(0)$ for $w_j$:

$$
\| \partial_{tt} v_m(0) \|_{L^2(\Omega)}^2 = \langle f(0), \partial_{tt} v_m(0) \rangle + \langle \Delta u_0, \partial_{tt} v_m(0) \rangle . \quad (2-16)
$$

We have used the fact that $v_m(0) = \partial_t v_m(0) = 0$. Now (2-16) implies

$$
\| \partial_{tt} v_m(0) \|_{L^2(\Omega)} \leq M, \quad (2-17)
$$

$M$ being a positive constant independent of $m$. Combining (2-13), (2-15) and (2-17), we derive

$$
\| \partial_{tt} v_m \|_{L^2(\Omega)} \leq M, \text{ for all } t \in [0, T], \quad (2-18)
$$

$$
\| \partial_t v_m \|_{H_1^1(\Omega)} \leq M, \text{ for all } t \in [0, T], \quad (2-19)
$$

where $M$ denotes positive constants independent of $m$. By means of (2-11), (2-18) and (2-19), we
can extract a subsequence still denoted by $\{v_m\}$ such that

$$
v_m \to v \text{ weak } * \text{ in } L^\infty(0, T; H_1^1(\Omega)) \quad (2-20)
$$

$$
\partial_t v_m \to \partial_t v \text{ weak } * \text{ in } L^\infty(0, T; H_1^1(\Omega)) \quad (2-21)
$$

$$
\partial_{tt} v_m \to \partial_{tt} v \text{ weak } * \text{ in } L^\infty(0, T; L^2(\Omega)) \quad (2-22)
$$

for some function $v$. By virtue of Lemma 1.4, we have

$$
(v_m - \phi)^- \to (v - \phi)^- \text{ in } C([0, T]; L^2(\partial\Omega)). \quad (2-23)
$$

We then make use of (2-20) through (2-23) to derive (2-4) and (2-5) through a standard procedure.
Remark 2.2. The purpose of the term $\frac{1}{k}\partial_t u_k$ in (2-3) is to obtain approximate solutions which are regular enough to justify manipulations in using a multiplier technique in the next subsection. The utility of that term is seen in (2-15).

## 2.2 A priori estimates independent of $k$.

We still suppress the subscript $k$ and let $u(x, t)$ be a solution constructed in Proposition 2.1 for a fixed $k > 0$. We can substitute $\partial_t u$ for $w$ in (2-3):

$$ \frac{1}{2} \frac{d}{dt} \| \partial_t u \|_{L^2(\Omega)}^2 + \frac{1}{2} \frac{d}{dt} a(u, u) = < f, \partial_t u > \\ - \frac{1}{2} k \frac{d}{dt} \int_{\Gamma_2} \{(u - \phi)^{-} \}^2 d\sigma - \frac{1}{k} \int_{\Gamma_2} (\partial_t u)^2 d\sigma, \quad (2-24) $$

for almost all $t \in (0, T)$. We now derive from (2-24)

$$ \| \partial_t u \|_{L^2(\Omega)} \le M, \text{ for all } t \in [0, T] \quad (2-25) $$

$$ \|u\|_{H_1^1(\Omega)} \le M, \text{ for all } t \in [0, T] \quad (2-26) $$

$$ \|(u - \phi)^{-} \|_{L^2(\Gamma_2)} \le M/\sqrt{k}, \text{ for all } t \in [0, T], \quad (2-27) $$

$$ \int_{0}^{T} \| \partial_t u \|_{L^2(\Gamma_2)}^{2} dt \le M k, \quad (2-28) $$

where $M$ denotes positive constants which are independent of $k$, and depend only on the number $L$ such that $\|u_0\|_{H_0^1(\Omega)} + \|u_1\|_{L^2(\Omega)} + \|f\|_{L^2(\Omega \times (0,T))} \le L$.

Next we shall adapt a multiplier technique to derive some uniform estimates in a neighborhood of $\partial\Omega$. Let us set

$$ \Omega_{\delta} = \{z \in \Omega : \mathrm{dist}(z, \partial\Omega) > \delta\} $$

for $0 < \delta \le \delta_0$. Obviously, $\partial\Omega_\delta = S_\delta$, which was defined in Section 1. We need to observe that solution $u$ of (2-1), (2-2) and (2-3) has sufficient regularity to justify manipulations in using a multiplier technique. First of all, (2-3) implies that

$$ \partial_{tt} u - \Delta u = f \quad (2-29) $$

holds in $D'(\Omega \times (0, T))$. By virtue of (2-1) and the assumption that $f \in C^1([0, T]; L^2(\Omega))$, we infer that

$$ \Delta u \in L^\infty(0, T; L^2(\Omega)). \quad (2-30) $$

In fact, (2-29) holds in $L^\infty(0, T; L^2(\Omega))$. Now (2-30) implies

$$ u \in L^\infty(0, T; H^2(\Omega_\delta)), \quad (2-31) $$
for each $\delta > 0$. Because of the mixed boundary condition on $\partial\Omega$, one cannot claim that $u \in L^\infty(0, T; H^2(\Omega))$. We also note that (2-1) implies

$$ u \in C([0, T]; H_1^1(\Omega)), \partial_t u \in C([0, T]; L^2(\Omega)), \quad (2.32) $$

possibly after a modification on a set of measure zero in t.

We now proceed to obtain the estimates. Using the function $h(x)$ constructed in Lemma 1.1, we multiply both sides of (2-29) by $(h \cdot \nabla)u$ and integrate over $\Omega_\delta \times (0, T)$, $0 < \delta \le \delta_0/2$:

$$ \begin{aligned} & \int_0^T \int_{\Omega_\delta} (\partial_t u)((h \cdot \nabla)u)dxdt \\ &= \int_{\Omega_\delta} (\partial_t u(T))((h \cdot \nabla)u(T))dx - \int_{\Omega_\delta} u_1((h \cdot \nabla)u_0)dx \\ & \quad - \int_0^T \int_{\Omega_\delta} (\partial_t u)((h \cdot \nabla)\partial_t u)dxdt = \end{aligned} $$

(applying the divergence theorem to the last integral)

$$ \begin{aligned} & = \int_{\Omega_\delta} (\partial_t u(T))((h \cdot \nabla)u(T))dx - \int_{\Omega_\delta} u_1((h \cdot \nabla)u_0)dx \\ & \quad - \frac{1}{2} \int_0^T \int_{S_\delta} (\partial_t u)^2 d\sigma_\delta dt + \frac{1}{2} \int_0^T \int_{\Omega_\delta} (\nabla \cdot h)(\partial_t u)^2 dxdt. \end{aligned} \quad (2.33) $$

$$ \begin{aligned} & \int_0^T \int_{\Omega_\delta} (-\Delta u)((h \cdot \nabla)u)dxdt \\ &= - \int_0^T \int_{S_\delta} \left(\frac{\partial u}{\partial \nu}\right)^2 d\sigma_\delta dt + \int_0^T \int_{\Omega_\delta} \nabla u \cdot \nabla \{(h \cdot \nabla)u\}dxdt \\ &= - \int_0^T \int_{S_\delta} \left(\frac{\partial u}{\partial \nu}\right)^2 d\sigma_\delta dt + \sum_{i,j=1}^n \int_0^T \int_{\Omega_\delta} \frac{\partial u}{\partial x_i} \frac{\partial h_j}{\partial x_i} \frac{\partial u}{\partial x_j}dxdt \\ & \quad - \int_0^T \int_{\Omega_\delta} \frac{1}{2}(\nabla \cdot h)|\nabla u|^2 dxdt + \frac{1}{2} \int_0^T \int_{S_\delta} |\nabla u|^2 d\sigma_\delta dt, \end{aligned} \quad (2.34) $$

where $\frac{\partial}{\partial \nu}$ is the unit outward normal derivative on $S_\delta$, i.e., $\frac{\partial}{\partial \nu} = h \cdot \nabla$ on $S_\delta$. Combining (2-33) and (2-34), we get

$$ \begin{aligned} & \frac{1}{2} \int_0^T \int_{S_\delta} (\partial_t u)^2 d\sigma_\delta dt - \frac{1}{2} \int_0^T \int_{S_\delta} |\nabla u|^2 d\sigma_\delta dt \\ & \quad + \int_0^T \int_{S_\delta} \left(\frac{\partial u}{\partial \nu}\right)^2 d\sigma_\delta dt = C(\delta), \end{aligned} \quad (2.35) $$

where

$$ \begin{aligned} C(\delta) = & - \int_0^T \int_{\Omega_\delta} f((h \cdot \nabla)u)dxdt \\ & + \text{obvious terms in (2.33) and (2.34)}. \end{aligned} \quad (2.36) $$
Recalling (2.25) and (2.26), it is evident that

$$|C(\delta)| \le M, \text{ for all } 0 < \delta \le \delta_0/2, \quad (2.37)$$

where $M$ is a positive constant which is independent of $k$ and depends only on the number $L$ such that

$$\|u_0\|_{H_0^1(\Omega)} + \|u_1\|_{L^2(\Omega)} + \|f\|_{L^2(\Omega \times (0,T))} \le L.$$

We next choose any $\rho(s) \in C(R)$ such that for all $s, 0 \le |\rho(s)| \le 1$, and set $\psi(x) = \rho(\text{dist}(x, \partial\Omega))$. Then, $\psi(x)$ is a continuous function and is constant on each $S_\delta, 0 < \delta \le \delta_0$. Hence, it follows from (2.35) that

$$\frac{1}{2} \int_0^T \int_0^{\delta^*} \int_{S_\delta} \psi(x) (\partial_t u)^2 d\sigma_\delta d\delta dt \\ - \frac{1}{2} \int_0^T \int_0^{\delta^*} \int_{S_\delta} \psi(x) |\nabla u|^2 d\sigma_\delta d\delta dt \\ + \int_0^T \int_0^{\delta^*} \int_{S_\delta} \psi(x) ((h \cdot \nabla)u)^2 d\sigma_\delta d\delta dt =$$

(using Lemma 1.2)

$$= \frac{1}{2} \int_0^T \int_{U_{\delta^*}} \psi(x) (\partial_t u)^2 dxdt - \frac{1}{2} \int_0^T \int_{U_{\delta^*}} \psi(x) |\nabla u|^2 dxdt \\ + \int_0^T \int_{U_{\delta^*}} \psi(x) ((h \cdot \nabla)u)^2 dxdt = 0(\delta^*), \quad (2.38)$$

for each $0 < \delta^* \le \delta_0/2$, where $0(\delta^*)$ is a number such that

$$|0(\delta^*)| \le M \delta^*, \quad (2.39)$$

where $M$ is the same positive number as in (2.37). (2.38) will be used in the next section.

### 3. Convergence of approximate solutions.

We assume that $u_0 \in H_0^1(\Omega)$, $u_1 \in L^2(\Omega)$, $f \in L^2(\Omega \times (0, T))$, $\phi \in C^1(\bar{\Omega})$ with $\phi \le 0$ on $\Gamma_2$, and define a solution of (0-1) through (0-5) by

**Definition 3.1** A function $u(x, t)$ is a solution of (0-1) through (0-5) if

$$u \in L^{\infty}(0, T; H_1^1(\Omega)), \partial_t u \in L^{\infty}(0, T; L^2(\Omega)) \cap C([0, T]; H^{-1/2}(\Omega)), \quad (3.1)$$

$$u(x, 0) = u_0(x), \partial_t u(x, 0) = u_1(x), \quad (3.2)$$

$$u(t) \in G, \text{ for almost all } t \in (0, T) \quad (3.3)$$

$$< \partial_t u(T), w(T) - u(T) > - < u_1, w(0) - u_0 >$$
$$ 
\begin{aligned}
& - \int_0^T < \partial_t u, \partial_t w - \partial_t u > dt + \int_0^T a(u, w - u) dt \\
& \geq \int_0^T < f, w - u > dt,
\end{aligned} 
\quad (3-4) 
$$

for every $w$ such that $w \in L^\infty(0, T; H_1^1(\Omega))$, $\partial_t w \in L^\infty(0, T; L^2(\Omega))$, $w(t) \in G$, for almost all $t \in (0, T)$.

We shall justify this definition. Suppose that $u$ is a solution according to the above definition. By taking $w = u \pm \xi$, $\xi \in C_0^\infty(\Omega \times (0, T))$, in (3-4), we find that (0-1) is satisfied in $D'(\Omega \times (0, T))$. (0-3) and (0-4) are satisfied in the sense $u(t) \in G$, for almost all $t \in (0, T)$. Now we assume further regularity on $u$:

$$ \partial_{tt} u \in L^2(0, T; L^2(\Omega \setminus E_\delta)) \quad (3-5) $$

$$ u \in L^2(0, T; H^2(\Omega \setminus E_\delta)) \quad (3-6) $$

for each $\delta > 0$, where we set

$$ E_\delta = \{x \in \bar{\Omega} : \text{dist}(x, \partial\Gamma_1) \leq \delta\}. $$

Recall that $\partial\Gamma_1 = \partial\Gamma_2$ is an $(n-2)$-dimensional, $C^2$-manifold by assumption. Then, (0-1) holds in a stronger sense:

$$ \int_0^T < \partial_{tt} u, w > dt - \int_0^T < \Delta u, w > dt = \int_0^T < f, w > dt \quad (3-7) $$

holds for every $w \in L^2(\Omega \times (0, T))$ such that $\text{supp } w(x, t) \cap E_\delta$ is empty, for almost all $t$, for some $\delta > 0$.

Next choose any nonnegative function $\xi(x) \in C_0^1(\Gamma_2)$. Then, we can extend $\xi(x)$ so that $\xi(x) \in C^1(\bar{\Omega})$ and $\text{supp } \xi \cap \{E_\delta \cup \Gamma_1\}$ is empty for some $\delta > 0$. Choose any nonnegative function $\eta(t) \in C_0^1((0, T))$ and take $w = u + \xi(x)\eta(t)$ in (3-4). By means of integration by parts and the divergence theorem (this procedure is justified by (3-5) and (3-6)), we find

$$ 
\begin{aligned}
& \int_0^T < \partial_{tt} u, \xi(x) \eta(t) > dt - \int_0^T < \Delta u, \xi(x) \eta(t) > dt \\
& + \int_0^T \int_{\Gamma_2} \left( \frac{\partial u}{\partial \nu} \right) \xi(x) \eta(t) d\sigma dt \geq \int_0^T < f, \xi(x) \eta(t) > dt.
\end{aligned} 
\quad (3-8) 
$$

It follows from (3-7) and (3-8) that

$$ \int_0^T \int_{\Gamma_2} \left( \frac{\partial u}{\partial \nu} \right) \xi(x) \eta(t) d\sigma dt \geq 0, \quad (3-9) $$

which is true for all nonnegative functions $\xi(x) \in C_0^1(\Gamma_2)$ and $\eta(t) \in C_0^1((0, T))$. Hence we can conclude that
$$ \frac{\partial u}{\partial \nu} \geq 0, \text{ almost everywhere in } \Gamma_2 \times (0, T). \qquad (3-10) $$

It remains to show the second part of (0-5). Let us choose a function $\xi_{\varepsilon}(x) \in C^1(\bar{\Omega})$ such that $0 \le \xi_{\varepsilon}(x) \le 1$, for all $x \in \bar{\Omega}$, $\xi_{\varepsilon} \equiv 1$ on $\Gamma_2 \setminus E_{\varepsilon}$ and $\text{supp } \xi_{\varepsilon} \cap \{E_{\varepsilon/2} \cup \Gamma_1\}$ is empty. We also choose any $\eta(t) \in C_0^1((0,T))$ such that $-1 \le \eta(t) \le 1$ for all $t$. Then, we set

$$ w = u + \xi_{\varepsilon}(x)\eta(t)(u - \phi) $$

in (3.4). As above, we use integration by parts, the divergence theorem and (3-7) to arrive at

$$ \int_0^T \int_{\Gamma_2} \xi_{\varepsilon}(x) \eta(t) \left( \frac{\partial u}{\partial \nu} \right) (u - \phi) d\sigma dt = 0. \qquad (3-11) $$

Here, we have the equality, because $-\eta(t)$ can be also chosen in place of $\eta(t)$. Since $u - \phi \ge 0$ almost everywhere in $\Gamma_2 \times (0, T)$ and $\delta > 0$ is arbitrary, (3-10) and (3-11) imply

$$ \left(\frac{\partial u}{\partial \nu}\right) (u - \phi) = 0, \text{ almost everywhere in } \Gamma_2 \times (0, T). \qquad (3-12) $$

This completes the justification of the above definition of a solution.

We now proceed to present the main result.

**THEOREM 3.2.** For $u_0 \in H_0^1(\Omega)$, $u_1 \in L^2(\Omega)$ and $f \in L^2(\Omega \times (0, T))$, there is a solution of (0-1) through (0-5).

**PROOF:** Choose sequences $\{u_{0k}\}$, $\{u_{1k}\}$ in $C_0^\infty(\Omega)$ and $f_k$ in $C^1([0, T]; L^2(\Omega))$ such that as $k \to \infty$,

$$ u_{0k} \to u_0 \text{ in } H_0^1(\Omega), \qquad (3-13) $$

$$ u_{1k} \to u_1 \text{ in } L^2(\Omega), \qquad (3-14) $$

$$ f_k \to f \text{ in } L^2(\Omega \times (0, T)). \qquad (3-15) $$

Using these, for each $k$, let $u_k$ be a solution constructed in Proposition 2.1. Then, each $u_k$ satisfies

$$ \begin{aligned} & \int_0^T \langle \partial_{tt} u_k, w - u_k \rangle dt + \int_0^T a(u_k, w - u_k) dt \\ &= \int_0^T \langle f, w - u_k \rangle dt + k \int_0^T \int_{\Gamma_2} (u_k - \phi)^{-} (w - u_k) d\sigma dt \\ & \quad - \frac{1}{k} \int_0^T \int_{\Gamma_2} (\partial_t u_k) (w - u_k) d\sigma dt, \end{aligned} \qquad (3-16) $$

for every $w(x, t)$ such that $w \in L^\infty(0, T; H_{\Gamma_1}^1(\Omega))$, $\partial_t w \in L^\infty(0, T; L^2(\Omega))$ and $w(t) \in G$, for almost all $t \in (0, T)$. Using integration by parts and

$$ \begin{aligned} & \int_0^T \int_{\Gamma_2} (u_k - \phi)^{-} (w - u_k) d\sigma dt \\ &= \int_0^T \int_{\Gamma_2} (u_k - \phi)^{-} (w - \phi) d\sigma dt - \int_0^T \int_{\Gamma_2} (u_k - \phi)^{-} (u_k - \phi) d\sigma dt \\ &\ge 0, \end{aligned} \qquad (3-17) $$
we obtain

$$
\begin{align*}
& < \partial_t u_k(T), w(T) - u_k(T) > - < u_{1k}, w(0) - u_{0k} > \\
& - \int_0^T < \partial_t u_k, \partial_t w > dt + \int_0^T \int_\Omega (\partial_t u_k)^2 dx dt \\
& + \int_0^T a(u_k, w) dt - \int_0^T \int_\Omega |\nabla u_k|^2 dx dt \\
& \geq \int_0^T < f, w - u_k > dt - \frac{1}{k} \int_0^T \int_{\Gamma_2} (\partial_t u_k) w d\sigma dt \\
& + \frac{1}{2k} \int_{\Gamma_2} (u_k(T))^2 d\sigma. \tag{3-18}
\end{align*}
$$

By virtue of (2-25) and (2-26), we can extract a subsequence still denoted by {$u_k$} such that as
$k \to \infty$,

$$
u_k \to u \text{ weak } * \text{ in } L^\infty(0, T; H_1^1(\Omega)), \quad (3-19)
$$

$$
\partial_t u_k \to \partial_t u \text{ weak } * \text{ in } L^\infty(0, T; L^2(\Omega)). \quad (3-20)
$$

Since each $u_k$ satisfies (2-29), we infer from (3-15) and (3-19) that

$$
\partial_{tt} u_k \to \partial_{tt} u \text{ weakly in } L^2(0, T; H^{-1}(\Omega)). \quad (3-21)
$$

Hence, by Lemma 1.4, we find that

$$
\partial_t u_k \to \partial_t u \text{ in } C([0, T]; H^{-\frac{1}{2}}(\Omega)), \quad (3-22)
$$

$$
u_k \to u \text{ in } C([0, T]; H^{\frac{1}{2}}(\Omega)), \quad (3-23)
$$

so that

$$
\langle \partial_t u_k(T), w(T) - u_k(T) \rangle \to \langle \partial_t u(T), w(T) - u(T) \rangle \quad (3-24)
$$

as $k \to \infty$, for each $w$. In the meantime, we notice that

$$
\begin{align*}
& \frac{1}{k} \left| \int_0^T \int_{\Gamma_2} (\partial_t u_k) w d\sigma dt \right| \\
& \le \frac{1}{k} \left( \int_0^T \int_{\Gamma_2} (\partial_t u_k)^2 d\sigma dt \right)^{\frac{1}{2}} \left( \int_0^T \int_{\Gamma_2} w^2 d\sigma dt \right)^{\frac{1}{2}}, \\
& \quad \text{(using (2-28))}, \\
& \le M \left( \int_0^T \int_{\Gamma_2} w^2 d\sigma dt \right)^{\frac{1}{2}} / \sqrt{k} \to 0, \tag{3-25}
\end{align*}
$$

as $k \to \infty$, for each $w$. Next we shall show that

$$
\int_0^T \int_\Omega ((\partial_t u)^2 - |\nabla u|^2) dx dt \geq \lim_{k \to \infty} \sup \int_0^T \int_\Omega ((\partial_t u_k)^2 - |\nabla u_k|^2) dx dt \quad (3-26)
$$
It is convenient to introduce the notation:

$$A_k = \frac{1}{2}(\partial_t u_k)^2 - \frac{1}{2}|\nabla u_k|^2 + ((h \cdot \nabla)u_k)^2,$$

$$B_k = ((h \cdot \nabla)u_k)^2,$$

$$A = \frac{1}{2}(\partial_t u)^2 - \frac{1}{2}|\nabla u|^2 + ((h \cdot \nabla)u)^2,$$

$$B = ((h \cdot \nabla)u)^2.$$

First of all, by Lemma 1.3, as $k \to \infty$,

$$A_k - B_k \to A - B \text{ in } D'(\Omega \times (0, T)). \quad (3-27)$$

Since the sequence {$A_k - B_k$} is bounded in $L^\infty(0, T; L^1(\Omega))$, we observe that as $k \to \infty$,

$$\int_0^T \int_{\Omega} \psi(x)(A_k - B_k)dxdt \to \int_0^T \int_{\Omega} \psi(x)(A - B)dxdt \quad (3-28)$$

for each $\psi(x) \in C_0(\Omega)$. Let us define a function $\rho_\delta(s) \in C(R)$ such that $\rho_\delta(s)$ is nondecreasing in $s$ and

$$\rho_\delta(s) = \begin{cases} 0 & \text{for all } s \le \delta/2 \\ 1 & \text{for all } s \ge \delta. \end{cases} \quad (3-29)$$

We then set $\psi_\delta(x) = \rho_\delta(\text{dist}(x, \partial\Omega))$. Fix any $0 < \delta \le \delta_0/2$. According to (3-28),

$$\int_0^T \int_{\Omega} \psi_\delta(x)(A_k - B_k)dxdt \to \int_0^T \int_{\Omega} \psi_\delta(x)(A - B)dxdt \quad (3-30)$$

as $k \to \infty$. Furthermore, with the aid of (2-38), we notice that

$$\int_0^T \int_{\Omega} (\psi_\delta(x) - 1) A_k dxdt = \int_0^T \int_{U_\delta} (\psi_\delta(x) - 1) A_k dxdt = 0_1(\delta), \quad (3-31)$$

where we can estimate $0_1(\delta)$ by using the number $M$ in (2-37), and (3-13), (3-14) and (3-15),

$$|0_1(\delta)| \le M\delta, \text{ for all } k. \quad (3-32)$$

Meanwhile we also have

$$\int_0^T \int_{\Omega} B_k dxdt \ge \int_0^T \int_{\Omega} \psi_\delta(x) B_k dxdt, \quad (3-33)$$

for each $k$, since $B_k$ is nonnegative and integrable, and $0 \le \psi_\delta(x) \le 1$, for all $x$. Since $A - B \in L^\infty(0, T; L^1(\Omega))$, it is apparent that

$$\int_0^T \int_{\Omega} (1 - \psi_\delta(x))(A - B)dxdt = \int_0^T \int_{U_\delta} (1 - \psi_\delta(x))(A - B)dxdt \\ = 0_2(\delta) \to 0 \text{ as } \delta \to 0. \quad (3-34)$$

Choose any $\epsilon > 0$. Then, by (3-32) and (3-34), there is $0 < \tilde{\delta}(\epsilon) \le \delta_0/2$ such that

$$|0_1(\delta)| + |0_2(\delta)| \le \epsilon, \quad (3-35)$$
for all $0 < \delta \le \tilde{\delta}(\epsilon)$ and all $k$. Using (3-30) and (3-34), we have for each $0 < \delta \le \delta_0/2$,

$$ 
\begin{aligned}
\int_0^T \int_\Omega (A-B)dxdt &= \int_0^T \int_\Omega \psi_\delta(x)(A-B)dxdt + o_2(\delta) \\
&= o_2(\delta) + \lim_{k \to \infty} \int_0^T \int_\Omega \psi_\delta(x)(A_k - B_k)dxdt \\
&= o_2(\delta) + \lim_{k \to \infty} \left\{ \int_0^T \int_\Omega (A_k - B_k)dxdt \right. \\
& \qquad \left. + \int_0^T \int_\Omega (\psi_\delta(x)-1)A_kdxdt - \int_0^T \int_\Omega \psi_\delta(x)B_kdxdt \right. \\
& \qquad \left. + \int_0^T \int_\Omega B_kdxdt \right\}. \qquad (3-36)
\end{aligned}
$$

In particular, for $0 < \delta \le \tilde{\delta}(\epsilon)$, we use (3-31), (3-33) and (3-35) to arrive at

$$ \int_0^T \int_\Omega (A-B)dxdt \ge -\epsilon + \limsup_{k \to \infty} \int_0^T \int_\Omega (A_k - B_k)dxdt. \qquad (3.37) $$

Since $\epsilon$ is arbitrary, it follows that

$$ \int_0^T \int_\Omega (A-B)dxdt \ge \limsup_{k \to \infty} \int_0^T \int_\Omega (A_k - B_k)dxdt, \qquad (3-38) $$

which shows (3-26). Combining (3-18), (3-19), (3-20), (3-24), (3-25) and (3-26), we conclude that (3-4) is satisfied. Finally, (3.23) implies

$$ (u_k - \phi)^- \to (u - \phi)^- \text{ in } C([0,T]; L^2(\Gamma_2)). \qquad (3-39) $$

But, each $u_k$ satisfies (2-27) and thus

$$ u \ge \phi, \text{ almost everywhere in } \Gamma_2, \text{ for each } t, \qquad (3-40) $$

from which (3-3) follows. Now the proof of Theorem 3.2 is complete.

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