Asymptotic analysis for the eikonal equation with the dynamical boundary conditions

We study the dynamical boundary value problem for Hamilton‐Jacobi equations of the eikonal type with a small parameter. We establish two results concerning the asymptotic behavior of solutions of the Hamilton‐Jacobi equations: one concerns with the convergence of solutions as the parameter goes to zero and the other with the large‐time asymptotics of solutions of the limit equation.


Introduction and the main results
We consider the initial-boundary value problem for the eikonal equation (1.1) Here ε ∈ (0, 1) is a parameter, Ω is a bounded open connected subset of R n , with C 1 boundary, Q := Ω × (0, ∞), ν(x) denotes the outer unit normal of Ω at x ∈ ∂Ω, and u 0 represents the initial data. We adapt the notion of viscosity solution as the notion of solution of eikonal equations in this article. Throughout this article we assume for simplicity that u 0 ∈ Lip(Ω), i.e., u 0 is Lipschitz continuous on Ω.
From this observation, we see that the above dynamical boundary condition is a kind of Neumann type boundary condition posed on the portion ∂Ω × (0, ∞) of the boundary ∂ Q of the domain Q.
Motivated with applications to superconductivity and surface evolution, Elliott-Giga-Goto [7] have studied the well-posedness of a Hamilton-Jacobi equation with a dynamical boundary condition, where the boundary condition is "tangential" to the lateral boundary and has the form u t (x, t) + g(x, t) = 0 (see also (1.4) below). As far as the authors know, a general study of Hamilton-Jacobi equations with dynamical boundary conditions goes back to Barles [3], where the well-posedness of dynamical boundary problems has been established (see for instance [3,Theoremé 4.11]).
We are also motivated by the recent studies on the Laplace equation x u(x, t) = 0 in Ω × (0, ∞) with the nonlinear dynamical boundary condition of the type u t (x, t) + ν(x) · D x u(x, t) = |u(x, t)| q , with a constant q > 1, due to Amann-Fila [1], Fila-Ishige-Kawakami [10] and others, where the blow-up phenomena and large time behavior of solutions are investigated. The Laplace equation above is, of course, the limit equation of the heat equations εu t (x, t) − x u(x, t) = 0 in Ω × (0, ∞) as ε → 0+. Here we replace these heat equations by the eikonal equations and the nonlinear dynamical boundary condition by the linear one as in (1.1).
We are thus concerned with the asymptotic behavior of the solution u ε of (1.1) as ε → 0+. Roughly speaking, if there is a limit function of u ε as ε → 0+, the limit function u should satisfy Regarding the initial condition for the limit function, as we will see in our main results, the solutions u ε develop an initial layer and the original initial condition u(·, 0) = u 0 does not make sense for the limit function u in general.
To overcome the difficulty of initial layer, we introduce a new (slower) time scale and, for the solutions u ε of (1.1), we set v ε (x, t) = u ε (x, εt) for (x, t) ∈ Q. Note that the function v ε satisfies (1. 3) In the informal level, by setting ε = 0 we get the problem for the limit function v of the v ε as ε → 0+: (1.4) The initial condition for the limit function u of the u ε is then given as the limit function v ∞ (x) of the solution v(x, t) of (1.4) as t → ∞. The recent developments concerning the large time asymptotics for solutions of Hamilton-Jacobi equations (see [4], [12]) suggest that the limit function v ∞ should be described as follows: define first the function v − 0 on Ω as the maximal subsolution of the stationary eikonal equation |Dv(x)| = 1 in Ω (1.5) among those v which satisfy v ≤ u 0 on Ω, and then v ∞ as the minimal solution of (1.5) among those v which satisfy v ≥ v − 0 on Ω. It is well-known (see the end of this section) that v − 0 and v ∞ are Lipschitz continuous on Ω with a Lipschitz bound depending only on the domain Ω. See [12,Lemma 2.2] for this Lipschitz continuity.
The main purpose of this paper is twofold. First, we consider the convergence of u ε as ε → 0+ and, second, we study the large time asymptotics for solutions of (1.2). Our result on the convergence of u ε is stated as follows: The stationary problem corresponding to (1.2) is the following.
|Du(x)| = 1 i nΩ, 1 + ν(x) · Du(x) = 0 on ∂Ω. (1.6) As we will see, this problem has a solution in C(Ω) and v ∞ is a supersolution of this problem. We define the function u ∞ as the maximal solution of (1.6) among those u which satisfy u ≤ v ∞ on Ω. Our result concerning the large time behavior of solutions of (1.2) is as follows.
Theorem 1.2 Let u ∈ C(Q) be a solution of (1.2) satisfying the initial condition u(·, 0) = v ∞ . Then We use the following notation: as above let Ω ⊂ R n be a bounded domain with C 1 boundary. By the implicit function theorem, there exists a function ρ ∈ C 1 (R n ) such that Ω = {x ∈ R n : ρ(x) < 0}, Note that Dρ(x) = |Dρ(x)|ν(x) for all x ∈ ∂Ω. We call such a function ρ a defining function of Ω. Let φ ∈ C(Ω) be a (viscosity) subsolution of |Dφ(x)| ≤ 1 in Ω. It is well-known (see [12,Proposition 1.14] and [2], [3], [14]) that this property is equivalent to the following Lipschitz property: |φ(x) − φ(y)| ≤ |x − y| for all x, y ∈ B and all ball B ⊂ Ω. Due to the C 1 regularity, connectedness and boundedness of Ω, any such function φ is Lipschitz continuous on Ω, with a uniform Lipschitz bound. (See [12,Lemma 2.2] for this.) In the following, the minimum of such uniform bounds will be denoted by L Ω . It is obvious that L Ω ≥ 1. For any bounded function f on a set A, f ∞,A denotes the sup-norm sup x∈A | f (x)|. For any T > 0, Q T denotes the domain Ω × (0, T ).

Preliminaries
We begin with the following theorem. Theorem 2.1 Let ε > 0. There exists a unique solution u ε ∈ Lip(Q) of (1.1).
Recall that u 0 ∈ Lip(Ω) is assumed here, which is crucial to conclude the Lipschitz continuity of u ε in the above theorem. On the other hand, for any continuous u 0 , one can show the unique existence of a uniformly continuous solution of (1.1). The above result is known in the literature (see for instance [3], [5]), but we give a proof here for the reader's convenience.
P r o o f . We first note that the uniqueness of solution of (1.1) is a direct consequence of Theorem A.1 (comparison theorem) in the appendix.
We next show that there exists a solution of (1.1) which is continuous on Q. Let L > 0 be a Lipschitz bound of It is easily checked that U + and U − are, respectively, a supersolution and a subsolution of According to Perron's method (see [6], [2], [3], [12] for instance), if we denote by S the set of all subsolutions φ of (2.1) such that U − ≤ φ ≤ U + on Q and set then u ε is a solution of (2.1) and u ε ∈ S. More precisely, u ε is a solution of (2.1) in the sense that u ε ∈ USC(Q), u ε is a subsolution of (2.1) and the lower semicontinuous envelope u ε * of u ε is a supersolution of (2.1). It is obvious that U − ≤ u ε * ≤ u ε ≤ U + on Q. We apply the comparison theorem (Theorem A.1 in the Appendix) to u ε and u ε * , to obtain u ε ≤ u ε * on Q. Thus we see that u ε = u ε * is continuous on Q. It is now obvious that u ε (x, 0) = u 0 (x) for all x ∈ Ω and that u ε is a solution of (1.1).
It is easy to check that u ε is a solution of (1.1) if and only if v ε is a solution of (1.3). Hence, Theorem 2.1 implies the following proposition.

Corollary 2.2 There exists a unique solution
where S 0 denotes the set of all subsolutions φ ∈ Lip(Ω) of (1.5) satisfying the inequality φ ≤ u 0 on Ω. It is a classical observation that S 0 = ∅ and the above formula gives a Lipschitz continuous subsolution of (1.5). The where S denotes the set of all solutions φ ∈ Lip(Ω) of (1.5) satisfying φ ≥ v − 0 on Ω. It is well-known (see also Proposition A.4 in the appendix or Perron's method as well) that S = ∅ and the above formula gives a solution in Lip(Ω) of (1.5).
The definition of u ∞ is related to the additive eigenvalue problem (or, ergodic problem): consider the problem of finding a pair (c, v) ∈ R × Lip(Ω) such that v is a solution of It is clear that if (c, v) is a solution of the above additive eigenvalue problem, so is the pair (c, v + A), with any constant A ∈ R. On the other hand, the following theorem assures that the additive eigenvalue c is unique and, indeed, c = 1.
2. We next show that c = 1 is the only possible choice for which (2.2) has a solution. We actually show that if there exist solutions (c 1 , u), (c 2 , v) ∈ R × Lip(Ω) of (2.2), then c 1 = c 2 . By symmetry, we only need to show that c 1 ≤ c 2 . To this end, we argue by contradiction. Thus, we assume that the inequality c 1 > c 2 holds. Let A > 0, and define the functions V, W ∈ Lip(Q) by It is easily seen that V and W are both solutions of (1.2). We select A sufficiently large so that But this is a contradiction since c 1 > c 2 . Thus we must have c 1 ≤ c 2 .

Proof of the main results
The existence and uniqueness of solution of (1.3) have been shown in Corollary 2.2. The following lemma is needed in our proof of the above theorem.

Lemma 3.2 (Comparison)
Let v ∈ Lip(Q) and w ∈ Lip(Q) be a subsolution and a supersolution of Then v ≤ w on Q.
The above comparison principle does not hold in general if the Lipschitz regularity of the functions v, w is removed. For this see Example A.5 in the Appendix.
P r o o f . Fix any ε > 0. Let M > 0 be a Lipschitz bound of the functions v and w. It is easily checked that the functions v ε (x, t) := v(x, t) − εM t and w ε (x, t) := w(x, t) + εM t are, respectively, a subsolution and a supersolution of Sending ε → 0 yields the desired inequality.
The following proposition is an immediate consequence of Theorem 3.1 and Lemma 3.2. Let 0 < ε < 1. Let M ≥ 1 be a Lipschitz bound of the function u 0 . It is easily checked that the functions U + , U − ∈ Lip(Q), given by U ± (x, t) = u 0 (x) ± Mt, are a supersolution and a subsolution of (1.3), respectively. By comparison (Theorem A.1), we get Hence, using Lemma A.2, we deduce that the collection {v ε } 0<ε<1 is equi-Lipschitz continuous on Q.
Thanks to the Ascoli-Arzela theorem, there are a sequence {ε j } ⊂ (0, 1) converging to zero and a function v ∈ Lip(Q) such that By the well-known stability property of viscosity solutions, we see that v is a solution of (1.4).
To complete the proof, we need to show that for any T > 0, For this, we argue by contradiction and suppose that there were a sequence {ε( j)} j ⊂ (0, 1) converging to zero and a constant 0 Passing to a subsequence and arguing as in the case of the sequence {ε j }, we may assume that there is a solution w ∈ Lip(Q) of (1.4) such that for all T > 0,

Theorem 3.4
Let v ∈ Lip(Q) be the solution of (1.4). Then Indeed, one can prove that there exists a constant T > 0 such that Fix an e ∈ R n so that |e| = 1. For any C ∈ R the function w C (x, t) := e · x + C is a solution of (3.1). Hence, choosing C > 0 so large that 2. We next show that for each x ∈ ∂Ω the function t → v(x, t) is nonincreasing on [0, ∞). We fix anyx ∈ ∂Ω and show that the function t → v(x, t) is nonincreasing on [0, ∞). To this end, we assume, by contradiction, that there were two positive numbers t 0 < t 1 such that v(x, t 0 ) < v(x, t 1 ).
For α ≥ 0 we introduce the function and all α > 0. Let α > 0 and let (x α , t α ) be a maximum point of the function α over Ω × [t 0 , t 1 ]. Note that This sequence of inequalities guarantees that lim α→∞ (x α , t α ) = (x,t). In particular, if α is sufficiently large, then t 0 < t α < t 1 . For such a large α, by the viscosity property of v, we have either Noting that lim α→∞ 2α(x α −x) − α 2 Dρ(x α ) = ∞ and sending α → ∞, we get ψ (t) ≤ 0, which contradicts our choice of ψ. Thus we see that for each x ∈ ∂Ω the function t → v(x, t) is nonincreasing on [0, ∞) and, therefore, the limit By the monotonicity of the function It is a standard observation (see [12,Proposition 1.10], [6], [2], [3] for instance) that v + and v − are a subsolution and a supersolution of (3.1), respectively. Because of the Lipschitz continuity and boundedness of v ± , we find that By the stability of the viscosity property under uniform convergence, we see that V + and V − are a subsolution and a supersolution of (3.1), respectively. It is easily seen that the functions V ± (x, t) in fact do not depend on t.
We may thus denote them respectively by V ± (x), and we have for all x ∈ Ω, Obviously we have for all x ∈ Ω, and the functions V + and V − are a subsolution and a supersolution of the eikonal equation (1.5), respectively. By the standard comparison result (see Lemma A.3 in the Appendix and also [2], [3], [14]), we get From these observations, we infer that for each t > 0 the function for all x ∈ Ω. By the constancy (3.5), it is now easy to check that v − is a subsolution of (3.1). Also, the function v ∞ (x), regarded as a function of (x, t), is a solution of (3.1). Hence, by comparison, Since v ∞ is a solution of (1.5), the function v ∞ is Lipschitz continuous on Ω with a Lipschitz bound L Ω . Hence, the functions v ∞ (x) ± L Ω t are a supersolution and a subsolution of (1.1), respectively. Using (3.7), by comparison (Theorem A.1), we get The functions u + and u − are called the half-relaxed limits of the functions u ε , and it is well-known (see [12,Theorem 1.3], [6], [2], [3] and u + and u − are a subsolution and a supersolution of (1.2), respectively. Due to estimate (3.8), we see that if we set u ± (x, 0) = v ∞ (x) for x ∈ Ω, then u + ∈ USC(Q) and u − ∈ LSC(Q). By comparison (Theorem A.1), we get u + ≤ u ≤ u − on Q, which shows that u + = u − = u on Q and moreover that as The function v ∞ is a solution of (1.5) and hence it is a supersolution of (1.6). Indeed, for any x ∈ ∂Ω and p ∈ D − v ∞ (x), we have two possibilities: either | p| ≥ 1 or | p| < 1, and if | p| < 1, , as a function of (x, t), is a supersolution of (3.9). By Note that the function u ∞ (x) is, as a function of (x, t), a solution of (3.9) and that Noting also that w is bounded and Lipschitz continuous on Q, we see that, as t → ∞, w(x, t) converges to w ∞ (x) uniformly on Ω for some function w ∞ ∈ Lip(Ω). Clearly, w ∞ is a solution of (1.6) and satisfies Because of the maximality of u ∞ , we conclude that w ∞ = u ∞ and that, as t → ∞, w(x, t) converges to u ∞ (x) uniformly on Ω.

Initial value problem for (1.2)
We discuss here the well-posedness of the initial value problem for (1.2) and consider first the initial value problem This problem has been studied in the previous sections, but it is overdetermined in its initial condition. Indeed, if u is a solution of (4.1) and continuous on Q, then the function u 0 should be a solution of |Du 0 (x)| = 1 in Ω and therefore it should be given by the boundary data u 0 | ∂Ω . This suggests another formulation: let u 0 ∈ C(∂Ω) and consider the initial value problem Hence, we may choose a sequence t j → 0+ such that the limit exists for all x ∈ Ω and the convergence is uniform on Ω. It is a standard observation that the limit function u 0 is a unique solution of the Dirichlet problem Moreover, it follows that, as t → 0+, u(x, t) converges toū 0 (x) uniformly for x ∈ Ω. We thus see that if ) is a solution of (4.2), then u is extended uniquely to a function on Q and the resulting function solves (4.1), with u 0 replaced by the unique solution v of the Dirichlet problem (4.3). It is well-known that there exists a solution v ∈ Lip(Ω) of (4.3) if and only if there exists a subsolution w ∈ Lip(Ω) of (4.3). Now, an existence result for (4.2) is stated as follows.
Before giving a proof of the above theorem, we present a comparison principle. To show the existence of a solution of (4.2), we consider problem (4.1), with u 0 replaced by the solution w ∈ Lip(Ω) of (4.3), and argue as in the proof of Theorem 2.1. Recall that the constant L Ω is a Lipschitz bound www.mn-journal.com of the function w. We define the functions U ± ∈ Lip(Q) by U ± (x, t) = w(x) ± L Ω t, and observe that U + and U − are a supersolution and a subsolution of (1.2), respectively. By Perron's method, we may find a function u ∈ USC(Q) such that u and the lower semicontinuous envelope u * are a subsolution and a supersolution of (1.2), respectively, and U − ≤ u * ≤ u ≤ U + on Q. By the comparison (see Theorem A.1) between u and u * , we see that u = u * ∈ C(Q). Also, by the comparison between the functions u(x, t + h) and u( Thus we see that u ∈ Lip(Q) and u is a solution of (4.1), with w in place of u 0 .

Variational formulas
We have studied several Hamilton-Jacobi equations of the eikonal type. We discuss in this section the variational (or optimal control) formulas for solutions of such Hamilton-Jacobi equations.
We recall first the general principle (the Bellman principle). Let U be an open subset of R m and Γ a closed subset of ∂U . Let γ : The Skorokhod problem has been investigated extensively in the literature (see [15]), and we refer to [12,Theorem 5.2], [13] for the existence results convenient for our discussion here. We denote by SP the set of all quadruples (X, l, τ, v) which satisfy (5.1). We consider the function on U , where V 0 , f and g are given functions on Γ, U × R m and ∂U , respectively, and the infimum is taken over all (X, l, τ, v) ∈ SP such that X (0) = x. This is an optimal control problem, where the function v plays the role of control and where the function V is called the value function. The dynamic programming principle leads to the boundary-value problem for the value function V : where V 0 is a given function representing the Dirichlet (or initial) data on Γ . We apply the above principle to find correct variational formulas for solutions of the Hamilton-Jacobi equations discussed in the previous sections.
We treat first Equation (1.1), where the Hamilton-Jacobi equation can be written as www.mn-journal.com vector field (ν(x), 1) on ∂Ω × (0, ∞), where ν(x) and ε are from (1.1), corresponds to the γ (x, t) in the above. Given functions (v, w) ∈ L ∞ ([0, ∞), R n × R), our current Skorokhod problem is stated as , R) are to be looked for. Accordingly, SP denotes the set of all sextuples (X, T, l, τ, v, w) satisfying (5.3) and the minimization problem at (x, t) ∈ Q, i.e., the value at (x, t) of the optimal control problem associated with (5.3) or (5.2) is stated as For any (X, T, l, τ, v, w) ∈ SP, with (X (0), T (0)) = (x, t) ∈ Q, if the integral is finite in the above minimization formula, then |v(t)| ≤ 1 and w(t) = −ε for a.e. t ∈ [0, τ] and τ is characterized by These observations suggest a modification of SP and we introduce SP(1.1; x, t) as the set of all triples (X, l, τ ) P r o o f . We write V (x, t) for the right-hand side of (5.5).
1. It is a standard observation that the dynamic programming principle holds: for any (x, t) ∈ Ω × (0, ∞) and δ ∈ (0, t), The above computation is easily justified by approximating u 0 by smooth functions, and we conclude that where V * denotes the lower semicontinuous envelope of the function V .
3. Next, we show that and observe that (X, l, τ ) ∈ SP(1.1; x, t) and This clearly shows that (5.9) holds. 4. We prove that V * is a subsolution of (1.1). Let φ ∈ C 1 (Q) and (x,t) ∈ Ω × (0, ∞), and assume that V * − φ has a strict maximum at (x,t). We treat here only the case wherex ∈ ∂Ω. The other case can be handled similarly and more easily. We argue by contradiction and hence assume that We choose a unit vector e ∈ R n so that |D x φ(x,t)| = −e · D x φ(x,t), and then a constant R ∈ (0,t) so that for In view of (5.7), we fix an r ∈ (0, R/2) so that ε −1 + 1 r < R/2, and choose a point ( Such a choice is possible since We set δ :=t − (t − r/2), and note that 0 <t − (t − r/2) < r and ε −1 + 1 δ < R/2. According to the existence results in [12], [13], there exists a solution (X, l, τ ) ∈ SP(1.1;x, δ) such thaṫ By (5.6), we get We may assume by adding a constant to the function φ that (V − φ)(x,t) = 0, which implies that Hence, we get By estimate (5.7) and our choice of δ, we have X (s) ∈ B R/2 (x) ⊂ B R (x). Now, using (5.10), we get which is a contradiction. Thus, V * is a subsolution of (1.1). 5. We next prove that V * is a supersolution of (1.1). The argument here is similar to that in the previous step. Let φ ∈ C 1 (Q) and (x,t) ∈ Ω × (0, ∞), and assume that V * − φ has a strict minimum at (x,t). Here again we treat only the case wherex ∈ ∂Ω. We assume by contradiction that We choose a constant R ∈ (0,t) so that for all ( As in the previous step, we may choose a point (x,t) ∈ Q and a constant δ > 0 such that We may assume as before that (V − φ)(x,t) = 0 and, hence, by (5.6), we may choose a triple (X, l, τ ) ∈ SP(1.1;x,t) so that V (x,t) + γ > τ + V (X (τ ),t − δ), which yields φ(x,t) > τ + φ(X (τ ),t − δ). Noting that X (s) ∈ B R (x) for all s ∈ we compute that which is a contradiction. We have thus shown that V * is a supersolution of (1.1). 6. We now apply a comparison result (for instance, Theorem A.1), to conclude that V * ≤ u ε ≤ V * on Q, which obviously shows that u ε = V on Q.
Replacing (l, t) by (εl, εt) in (5.4) is a simple modification to get the right Skorokhod problem for (1.3). We thus denote by SP(1.3; x, t) the set of all triples (X, l, (5.12) The following proposition is an immediate consequence of the previous theorem. The variational formula for the solution of (1.4) is stated as follows.

Theorem 5.3 Let v ∈ Lip(Q) be the solution of (1.4).
Then Because of the lack of a "good" comparison theorem (see Lemma 3.2 and Example A.5), our strategy for the proof of the above theorem differs substantially from that of Theorem 5.1.
P r o o f . We write V (x, t) for the right-hand side of (5.14). The dynamic programming principle holds, which is stated as www.mn-journal.com 2. Let A > 0 be a Lipschitz bound of the function u 0 . Following Step 2 of the proof of Theorem 5.1, we obtain V (x, t) ≥ u 0 (x) − At for all (x, t) ∈ Q. Hence, using (5.15), we get Thus, combining (5.16) and (  To see this, let x, y ∈ Ω and (X, l, τ ) ∈ SP(x) be such that X (τ ) = y. Let ρ be a defining function of Ω and note that for any t ∈ [0, τ ], if X (t) ∈ ∂Ω, then the function s → ρ(X (s)) attains the maximum value 0 at t. Hence, if t ∈ (0, τ ) is a point where X (t) ∈ ∂Ω and the function X is differentiable at t, then that is, two vectors ν(X (t)) andẊ (t) are perpendicular. Accordingly, we have Thus, the first condition in (5.18) is equivalent to condition (5.19). Before giving a proof of the above theorem, we make similar observations for v − 0 , v ∞ and u ∞ . We introduce two "distance" functions d Ω and λ Ω on Ω × Ω, where d Ω (x, y) is defined as the infimum of all positive numbers τ for www.mn-journal.com which there exists a function X ∈ Lip([0, τ], Ω) such that X (t) ∈ Ω for all t ∈ [0, τ], X (0) = x and X (τ ) = y, and λ Ω (x, y) is defined by where SP(x) denotes the set of all triples (X, l, τ ) Note that by (5.19) that λ (x, y) ≥ 0 for all x, y ∈ , and that We note as well that and a subsolution of (1.6).
P r o o f . 1. Note that the function v(x) = |x − y| in R n , with y ∈ R n , is a solution of |Dv(x)| = 1 in R n \ {y} and is a subsolution of |Dv(x)| = 1 in R n .
2. Let y ∈ Ω and let B ⊂ Ω be an open ball such that y ∈ B. According to the dynamic programming principle, we deduce that for any A consideration based on the dynamic programming principle similar to the above shows that if ε > 0 is sufficiently small, then, for all x ∈ ∂Ω and y ∈ Ω, This and the Lipschitz continuity of the functions x → d Ω (x, y) and x → λ Ω (x, y) in Ω guarantee that these functions are Lipschitz continuous on Ω.
3. By following the argument (Steps 4 and 5) of the proof of Theorem 5.1, it is now not hard to check that the function x → λ Ω (x, y) on Ω is a solution of (5.23).
(ii) If u ∈ Lip(Ω) is a solution of (1.6), then By Lemma 5.7, we have u(x) ≤ V (x) for all x ∈ Ω. By the definition of V , we see that V (x) ≤ u(x) for all x ∈ ∂Ω. Hence, we have u(x) = V (x) for all x ∈ ∂Ω. According to Proposition A.4, the function V is a solution of (1.5). Hence, by Lemma A.3, we conclude that u = V on Ω.
The proof of (ii) is similar to the above, and we skip it here. We show that as well as the locally boundedness of the function V . Once this is done, we just need to follow Steps 4, 5 and 6 of the proof of Theorem 5.1. 2. It is a standard observation that for each t > 0 the function w : x → u(x, t) is a solution of the eikonal equation |Dw(x)| = 1 in Ω. By assumption, we have u ∈ Lip(Q). Hence, by the stability of the viscosity property, we see that u 0 is a solution of |Dw(x)| = 1 in Ω. As in Step 2 of the proof of Theorem 5.1, we easily find a constant A > 0 such that V (x, t) ≥ u 0 (x) − At for all (x, t) ∈ Q, which proves that V is locally bounded below in Q and that V * (x, 0) ≥ u 0 (x) for all x ∈ Ω.
P r o o f o f T h e o r e m 5.5. 1. We write V (x) for the right-hand side of (5.20). Since v − 0 is a subsolution of (1.5), by Lemma 5.
On the other hand, in view of Proposition A.4 we see that V is a subsolution of (1.5). Also, we have Let V (x) denote the right-hand side of (5.21). By Lemma 5.7, we have y) for all x, y ∈ Ω, and hence, v − 0 ≤ V on Ω. In view of Proposition A.4, the function V is a solution of (1.5). By the minimality for all x ∈ ∂Ω. Now, by comparison (Lemma A.3), we get v ∞ = V on Ω. 3. Let V (x) denote the right-hand side of (5.22). As noted before, the function v ∞ is a supersolution of (1.6). In Step 2 above, we have observed that v ∞ = v − 0 on ∂Ω. According to Lemma 5.6, the function V is a solution of (1.6). Since V ≤ v ∞ on ∂Ω, by comparison, we get V ≤ v ∞ on Ω. Hence, by the maximality of u ∞ , we see that V ≤ u ∞ on Ω. On the other hand, by (ii) of Lemma 5.8, we find that Thus, we have u ∞ = V on Ω.

More on the function λ Ω
By the assumption that Ω is a bounded, open connected subset of R n and is of class C 1 , we deduce that ∂Ω consists of a finite number of connected components Γ i , with i = 1, 2, . . . , N .
We have shown in the proof of Theorem 2.3 that the function x → dist(x, ∂Ω) on Ω is a solution of (1.6). The same proof shows that for each i =
We divide the proof of Theorem 6.1 into two parts.
P r o o f o f T h e o r e m 6.1. P a r t 1. We fix any y ∈ Ω and write v(x) for the right-hand side of formula (6.2). Here we prove that We first prove that v is a solution of (1.6). Let i = 1, . . . , N . By the definition of a i (y), for any sequence (i 1 , . . . , i m ) ∈ I such that i m = i, we have In particular, if m = 1, then we get Also, for any j = 1, . . . , N , if we choose (i 1 , . . . , i m ) ∈ I , with i m−1 = j, optimally so that Hence, by the definition of v, we see that v(x) = a i (y) for all x ∈ Γ i . Note as well by the definition of v that v(x) ≤ a i (y) + dist(x, Γ i ) for all x ∈ Ω.
Let ε > 0 and set There exists an open neighborhood V ε , relative to R n , of It is now a standard observation that v ε is a solution of (6.1). It is clear that Hence, by the stability of the viscosity property under uniform convergence, we see that v is a solution of (6.1). Since our choice of i is arbitrary, we may conclude that v is a solution of (1.6). Noting that v(y) = 0, by Lemma 5.7, we conclude that (6.3) holds.
Recalling the Jordan-Brouwer separation theorem (see for instance [11]), since the Γ i are compact, connected C 1 hypersurfaces, we see that for each i = 1, 2, . . . , N the open subset R n \ Γ i of R n has exactly two connected components O + i and O − i . Since Ω is connected and does not intersect ∂Ω = ∪ i Γ i , for each i we have either  1]. Now, since Γ i is locally diffeomorphic to a hyperplane, it is not hard to find a small constant ε > 0 and a continuous curve Here it is assumed that 0 < σ − ε < τ + ε < 1. Moreover, we may select the curve η so that the distance of the curve η to the is as small as required. Consequently, we may assume that These together yield a contradiction: 1. Assume first that φ(t) ∈ Ω for all t ∈ [0, 1]. Fix any ε > 0, set τ ε := |x − y| + ε and φ ε (t) := φ τ −1 ε t for t ∈ [0, τ ε ], and note that (φ ε , τ ε , 0) ∈ SP(x) that is, l(t) ≡ 0 in the usual notation and φ(τ ε ) = y. By the definition of λ Ω , we get λ Ω (x, y) ≤ τ ε = |x − y| + ε, which shows that (6.4) holds in this case.
According to Lemma 6.2, we have λ Ω (φ(s k ), φ(t k )) = 0 for all k = 1, . . . , m. Noting that and arguing as in Step 1, we get Adding these all together, we obtain The proof is complete.
P r o o f o f T h e o r e m 6.1. P a r t 2. As in Part 1, we fix any y ∈ Ω and write v(x) for the right-hand side of formula (6.2). We show that which will complete the proof of the theorem. Fix any i ∈ {1, . . . , N }. There exists a sequence (i 1 , . . . , i m ) ∈ I such that i m = i and We may choose sequences (x 1 , . . . , x m ) ∈ (∂Ω) m and (y 1 , . . . , y m−1 ) ∈ (∂Ω) m−1 so that www.mn-journal.com By the triangle inequality, we get λ Ω (y k , x k+1 ).
We first consider the problem we may assume that max Ω×∂(0, T ) (u − v) ≤ 0. We now adapt the classical comparison argument for the Neumann type boundary value problem (see [13,Theorem 3.1]) for Hamilton-Jacobi equations to the present situation, to obtain a contradiction, which completes the proof.
The stationary eikonal equation (1.5) is of a special importance in this article and the following is a well-known comparison result (see [2], [3], [14] for instance) for (1.5).
The following proposition is a well-known result for convex Hamilton-Jacobi equations (see, for instance, [8], [9], [12] for the proof). Then the function u is a subsolution of (A.4).
As is well-known, if we replace "inf" by "sup" in the above definition of u, the same conclusion as above holds without the convexity of H . As Lemma 3.2 states, the comparison principle holds for Lipschitz continuous subsolutions and supersolution of the above problem. In what follows we show that the comparison principle does not hold for semicontinuous subsolutions and supersolutions of (3.1).
Note that v is a classical solution of (3.1) and that w is a classical solution of w t + |D x w| = 1 in Ω × (0, ∞) and a classical subsolution of (3.1). Accordingly, u is a viscosity subsolution of (3.1).
Our claim here is that u is a solution of (3.1) satisfying the initial condition (A.5). It is clear that u(x, 0) = 2x for all x ∈ Ω. We have already checked that u is a subsolution of (3.1). Note that u(x, t) = v(x) if t > x, which shows that u is a classical solution of u t + |D x u| = 1 in Q + , where Q + := {(x, t) : t > x}. Similarly, noting that u = w if t < x, we see that u is a classical solution of u t (x, t) + |D x u(x, t)| = 1 in Q − , where Q − := {(x, t) ∈ Q : t < x}. Fix any (x,t) ∈ Ω × (0, ∞) such thatt ≤x. Let φ ∈ C 1 Q) and assume that u − φ has a minimum at (x,t). The function r → (u − φ)(r +x −t, r ) = r + 2(x −t) − φ(r +x −t, r ) on (0,t] has a minimum at r =t and we get 0 ≥ 1 − φ x (x,t) − φ t (x,t), that is, φ t (x,t) + φ x (x,t) ≥ 1, which shows that φ t (x,t) + |φ x (x,t)| ≥ 1. This assures that u is a supersolution of (3.1). Thus, we conclude that u is a viscosity solution of (3.1).
We The functions u and U differ only on the set {1} × (0, ∞) and the function U is upper semicontinuous on Q. It is obvious to see that U is a viscosity subsolution of (3.1) and that U (x, 0) = 2x = u(x, 0) for all x ∈ [0, 1] = Ω. Moreover, the inequality U ≤ u on Q does not hold. That is, in the framework of semicontinuous viscosity solutions, the comparison principle does not hold.