ON A HYPERBOLIC-PARABOLIC MIXED TYPE EQUATION

. In this paper, the hyperbolic-parabolic mixed type equation ∂u ∂t = ∆ A ( u ) + div( b ( u )) , ( x,t ) ∈ Ω × (0 ,T ) , with the homogeneous boundary condition is considered. We ﬁnd that only a part of the boundary condition is able to ensure the posedness of the solutions. By introducing a new kind of entropy solution matching the part boundary condition in a special way, we obtain the existence of the solution by the BV estimate method, and establish the stability of the solutions by the Kruzkov bi-variables method.


Introduction. Consider the equation
with the homogeneous boundary condition, where Ω ⊂ R N is an open bounded domain with the appropriately smooth boundary ∂Ω. We assume that A(u) = u 0 a(s)ds, a(s) ≥ 0, a(0) = 0.
Equation (1) is of hyperbolic-parabolic mixed type. If A ≡ 0, it becomes the wellknown conservation law equation, which belongs to the hyperbolic equations. The equation of the flow in a porous medium, ∂u ∂t = ∆u m , and the stationary boundary layer theory equation are two special cases of equation (1), which belong to the degenerate parabolic equations. Here, in equation (2), v 0 (x) is an initial velocity, and ν is the viscous coefficient of the flow. One can refer to [22] for details. For the Cauchy problem of equation (1), there are considerable attention on its well-posedness, see [1,2,3,4,5,6,9,14,17,27,28,29,30,33,35,37,38] and the references therein.
While considering the initial-boundary value problem of equation (1), usually we need the initial condition u(x, 0) = u 0 (x), x ∈ Ω. (3) However, can we give the Dirichlet homogeneous boundary condition as follows?
When the equation is weakly degenerate, it is well known that we can give the Dirichlet homogeneous boundary condition (4) [33]. When the equation is strongly degenerate, there are two ways to deal with the corresponding problem. One is based on the BV analysis technique. In general, instead of the whole boundary ∂Ω, only a part of the boundary Σ p ⊆ ∂Ω on which the trace of u can be endowed in the traditional sense [31,32,34]: The other way is that, the entropy solutions are defined only in L ∞ space, the existence of the traditional trace (which was called the strong trace in [16]) on the boundary is not guaranteed, so the boundary condition is not directly shown in the traditional way as (4), it is implicitly contained in a family entropy inequalities [8,13,16,[18][19][20][21]26]. The advantage of the first way lies in that, we can figure out the portion of the boundary on which we can impose the boundary value, whereas the rest portion of the boundary is free from the limitation of the boundary condition.
If the domain Ω = R N + is the half space of R N , in our previous work [37], we studied the initial-boundary value problem of equation (1) in the half space We have proved that if b N (0) < 0, we can give the general Dirichlet boundary condition u(x, t) = 0, (x, t) ∈ ∂R N + × (0, T ), which is satisfied in a particular weak sense. But if b N (0) ≥ 0, then no boundary condition is necessary, the solution of the equation is free from any limitation of the boundary condition.
In this paper, we continue to investigate how to find a suitable homogeneous boundary condition as (5) by the first way. According to [37], if we regard equation (1) as a "linear" degenerate elliptic equation [10][11][23][24]), then we know that Σ p defined in (5) should be as In this study, we will introduce a new kind of entropy solution to match the above the homogeneous boundary condition (5). The aim of this paper is to get the well-posedness of the new kind of entropy solutions.
2. Main results. For small η > 0, let Obviously h η (s) ∈ C(R), and Here and in what follows, Now, we can consider two cases. Case I. If Σ 1 = ∅ is a real subset of ∂Ω, we will show that if equation (1) is degenerate on the boundary, instead of the usual Dirichlet homogeneous boundary condition (4), we only can give the part boundary condition as to get the well-posedness of the initial-boundary value problem of equation (1). In fact, by the definition of Σ 1ηk , we know that Let k → 0. We know that b i (0)n i (x) < 0, which is in accordance with (6).
Definition 2.1. If Σ 1 = ∅ is a subset of ∂Ω, a function u is said to be the entropy solution of equation (1) with the initial condition (3) and the homogeneous boundary condition (7), provided the following conditions are true.
1. u satisfies T ] , and suppϕ 2 , suppϕ 1 ⊂ Ω × (0, T ), with k ∈ R and small η > 0, u satisfies 3. The homogeneous boundary value is true in the sense of trace, 4. The initial value is true in the sense of Here the pairs of equal indices imply a summation from 1 up to N , and Case II. If Σ 1 = ∅, no any other boundary value condition is necessary. In other words, the solution of equation (1) is completely controlled by the initial condition. Then Definition 2.1 can be re-stated as Definition 2.2. A function u is said to be the entropy solution of equation (1) with the initial value (3), if the following three conditions are true.
1. u satisfies 2. For any ϕ ∈ C 2 0 (Q T ), ϕ ≥ 0, k ∈ R, and small η > 0, u satisfies 3. The initial value is true in the sense of On the one hand, if equation (1) has a classical solution u, multiplying (1) by ϕ 1 S η (u − k) and integrating over Q T , we are able to show that u satisfies Definition 2.1. Multiplying (1) by ϕS η (u − k) and integrating over Q T , we are able to show that u satisfies Definition 2.2.
On the other hand, in the first case of that Σ 1 = ∅ is a real subset of ∂Ω, let η → 0 in (8). One has Let ϕ 2 = 0 and so ϕ 1 | ∂Ω×(0,T ) = 0. Then In the second case of that Σ 1 = ∅, let η → 0 in (11). We have Thus if u is the entropy solution in Definition 2.1( Definition 2.2), then u is a entropy solution defined in [3,27,37] et al. (1) with the initial-boundary value conditions (3) and (7) has a entropy solution in the sense of Definition 2.1. If Σ 1 = ∅, then equation (1) with the initial condition (3) has a entropy solution in the sense of Definition 2.2.
We start the proof by considering the following regularized problem with We denote the solution of problem (12)-(14) as u ε . Let u be the solution of the problem (1)-(3)-(7). If it is the limit of the approximate solution u ε , we say u is a viscous solution. Since the solution u is in the BV sense, the trace of u on the boundary ∂Ω can always be defined in the traditional way. Now, if u is a viscous solution, we know that on the whole boundary ∂Ω, it has However, for a general entropy solution u in the sense of Definition 2.1, by (9), we know that it satisfies γu | Σ1×(0,T ) = 0. Whereas on the other part Σ 2 × (0, T ), it is unclear if u is equivalent to 0 or not. Based on this observation, we will prove the following stabilities.
respectively. Suppose that the distance function d(x) satisfies (17), and Then it has Clearly, if u, v appearing in Theorem 2.4-Theorem 2.5 are the two viscous solutions, then we have 3. Existence of the solution.
. We now consider the regularized problem (12)- (14). There is a classical solution u ε ∈ C 2 (Q T ) C 3 (Q T ) of this problem provided that A and b i satisfy the assumptions in Theorem 2.3, one can refer to the eighth chapter of [12] for details.
Firstly, since u 0 (x) ∈ L ∞ (Ω) is smooth, by the maximum principle, we have Secondly, let us make the BV estimates of u ε . To the end, we begin with the local coordinates of the boundary ∂Ω.
Let δ 0 > 0 be small enough such that where V τ is a region, on which one can introduce local coordinates y k = F k τ (x)(k = 1, 2, · · · , N ), y N | ∂Ω = 0, with F k τ appropriately smooth and F N τ = F N l , such that the y N −axes coincides with the inner normal vector.
Let u ε be the solution of the approximate problem (12)- (14). If the assumptions of Theorem 2.3 are true, then we have with constants c i (i = 1, 2) are independent of ε.
Theorem 3.3. Let u ε be the solution of the approximate problem (12)- (14). If the assumptions of Theorem 2.3 are true, then it has where c is independent of ε, and Proof. Differentiate (12) with respect to x s , s = 1, 2, · · · , N, N + 1, x N +1 = t, and sum up for s after multiplying the resulting relation by u εxs |graduε| . In what follows, we simply denote u ε by u, and ∂Ω by Σ. Integrating it over Ω yields As usual, pairs of the indices of s imply a summation from 1 to N + 1, pairs of the indices of i, j imply a summation from 1 to N , and where ξ s = u xs . So we have From (19)- (20), by the assumption a(0) = 0, we have By observing that on Σ, there holds then the surface integrals in (21) can be rewritten as We use the local coordinates on V τ , τ = 1, 2, · · · , n: y k = F k τ (x), k = 1, 2, · · · , N, y m | Σ = 0. By a direct computation [31] on Σ V τ , we have where F k = F k τ . By the fact of that the inner normal vector is By Lemma 3.2, we see that lim η→0 S can be estimated by |gradu| L1(Ω) . Thus, letting η → 0, and noticing that By Gronwall's inequality, we get Ω |gradu|dx ≤ c.
By (12) and (22), it is easy to see that Thus, there exists a subsequence {u εn } of u ε and a function u ∈ BV (Q T ) ∩L ∞ (Q T ) such that u εn → u a.e. on Q T .
Proof of Theorem 2.3. We now prove that u is a generalized solution of the problem (1)-(3)-(7). For any ϕ(x, t) ∈ C 1 0 (Q T ), it has The above equality is also true for any ϕ(x, t) ∈ L 2 (Q T ). By Hölder's inequality, from (23)
4. Proof of Theorem 2.4. Let Γ u be the set of all jump points of u ∈ BV (Q T ), v be the normal of Γ u at X = (x, t), u + (X) and u − (X) be the approximate limits of u at X ∈ Γ u with respect to (v, Y − X) > 0 and (v, Y − X) < 0 respectively. For the continuous function p(u, x, t) and u ∈ BV (Q T ), we define which is called the composite mean value of p. For a given t, we denote Γ t u , H t , (v t 1 , · · · , v t N ) and u t ± as all jump points of u(·, t), Housdorff measure of Γ t u , the unit normal vector of Γ t u , and the asymptotic limit of u(·, t) respectively. Moreover, if f (s) ∈ C 1 (R), u ∈ BV (Q T ), then f (u) ∈ BV (Q T ) and This lemma can be proved in a similar way as described [35], we omit the details. The proof of Theorem 2.4. Let u, v be two entropy solutions of equation (1) with initial values Let We choose k = v(y, τ ), l = u(x, t), ϕ 1 = ψ(x, t, y, τ ) in (33) and (34), integrate it over Q T . It gives Here ∆ x is the usual Laplacian operator corresponding to the variable x, and ∇ x is the gradient operator corresponding to the variable x.
Consequently, the desired result follows by letting s → 0.

Proof of Theorem 2.5.
Proof. If Σ 1 = ∅, let u and v be two entropy solutions of equation (1) in the sense of Definition 2.2. Suppose that the initial values are u(x, 0) = u 0 (x), v(x, 0) = v 0 (x), and the boundary values are the same as (18).
Consequently, the desired result follows by letting s → 0.