On the asymptotic of solution to the Dirichlet problem for hyperbolic equations in cylinders with edges

In this paper, we consider the Dirichlet problem for second-order hyperbolic equations whose coefficients depend on both time and spatial variables in a cylinder with edges. The asymptotic behaviour of the solution near the edge is studied.


Introduction
Let Ω be a bounded domain in R n , n ≥ 2, with the boundary ∂Ω consisting of two surfaces Γ 1 , Γ 2 which intersect along a manifold l 0 .Assume that in a neighbourhood of each point of l 0 the set Ω is diffeomorphic to a dihedral angle.Assume that in a neighbourhood of each point of l 0 the set Ω is diffeomorphic to a dihedral angle.For any P ∈ l 0 , two half-spaces T 1 (P), and T 2 (P) tangent to Ω, and a two-dimensional plane π(P) normal to l 0 are defined.We denote by ν(P) the angle in the plane π(P) (on the side of Ω) bounded by the rays R 1 = T 1 (P) ∩ π(P), R 2 = T 2 (P) ∩ π(P) and by β(P) the aperture of this angle.
In the cylinder Q = Ω × R, we study a class of second-order hyperbolic equations.The Dirichlet boundary condition is given on the boundary ∂Q = ∂Ω × R of the cylinder.Our goal is to describe the behaviour of the solutions near the edges.There are some approaches to this issue.For systems or equations dealt with in [13,6,12] whose coefficients are independent of the time variable, B. A. Plamenevssky used Fourier transform to reduce the problem to an elliptic one with a parameter.In contrast to [13] and [12], in this paper, we consider equations with coefficients depending on both of time and spatial variables.We develop the approach suggested in [2] to demonstrate the asymptotic representation of the solution of the problem mentioned above near the edges.Furthermore, we investigate the unique solvability of the problem and the regularity of solutions in weighted Sobolev spaces.
be a second-order partial differential operator, where a ij (x, t), b i (x, t) and c(x, t) are real-valued functions on Q belonging to C ∞ (Q).Moreover, suppose that a ij = a ji , i, j = 1, . . ., n, are continuous in x ∈ Ω uniformly with respect to t ∈ R and for all ξ ∈ R n \{0} and (x, t) ∈ Q, µ 0 = const > 0. In the present work, we consider the Dirichlet problem u| ∂Q = 0. (1.3) Let us introduce some functional spaces used in this paper.We denote by H l (Ω) and Hl (Ω) the usual Sobolev spaces as in [1].Let α ∈ R, we introduce the space H l α (Ω) as the weighted Sobolev space of all functions u defined on Ω with the norm we denote the weighted Sobolev spaces of functions u defined on Q with the norms Let us denote the time-dependent bilinear form.Applying condition (1.1) and similar arguments as the proof of the Gårding inequality, it follows that 3) if and only if u(x, t) ∈ H1,1 (Q, γ), and for any T > 0 the equality holds for all v ∈ H1,1 (Q, −γ), v(x, t) = 0, t ≥ T.
The problems for nonstationary systems or equations in nonsmooth domains also have been investigated in [2,3,4,5,10,11], in which the authors obtain results on the regularity of solutions in weighted Sobolev spaces and asymptotic behaviour of solutions in the neighbourhood of the conical points.However, the problems are considered in domains with conical points and with initial conditions.Different from the above-mentioned papers, we consider the problem without initial conditions in domains with edges.The paper is organized as follows.In Section 2, we present the results on the unique solvability of the problem.The regularity of the generalized solution is stated in Section 3. The main result, Theorem 4.2, is given in Section 4.

The unique solvability
In this section, we will establish the unique solvability and the regularity in time variable of the solution for problem (1.2)-(1.3).Furthermore, some energy estimates of the solution are proven.The solvability condition of problem (1.2)-(1.3)is: the right hand side f (external force) of (1.2) belongs to H 0 α (Q, γ) where γ is a sufficiently large positive number.This condition is more applicable than 3) has a generalized solution u in the space H1,1 (Q, γ) and where C is a constant independent of u and f .
To prove the theorem, we construct an approximate sequence u h of solution u of the problem (1.2)-(1.3).It is known that there is a smooth function χ(t) which is equal to 1 on [1, +∞), is equal to 0 on (−∞, 0] and assumes values from [0, 1] on [0, 1] (see [14,Thm. 5.5] for more details).Moreover, we can suppose that all derivatives of χ(t) are bounded.Let h ∈ (−∞, 0] be an integer.Set where , we consider the following problem in the cylinder Q h : ) Lemma 2.2.Suppose that the assumption of Theorem 2.1 is satisfied.For any h fixed, there exists a solution u h in the space H1,1 (Q h , γ) of the problem (2.2)-(2.4)and the following estimate holds where C is a constant independent of h.
Proof.We will prove the existence by Galerkin's approximating method.Let {ω k (x)} ∞ k=1 be an orthogonal basis of H1 (Ω) which is orthonormal in L 2 (Ω).Put where C N k (t), k = 1, . . ., N, is the solution of the following ordinary differential system: with the initial conditions Let us multiply (2.7) by C N kt (t), then take the sum with respect to k from 1 to N to arrive at Integrating the equality above from h to t we find that Let us evaluate the right-hand side of (2.9).Integrating by parts, we have By the Cauchy-Schwarz inequality and the Hardy inequality, for an arbitrary positive number α ∈ [0, 1], it follows from the equality above that where > 0 and C( ) is a constant independent of N, h.We consider the second term in the left-hand side of (2.9), we can write (2.11) It is easy to see that for the symmetric bilinear form The equality (2.12) implies and we also note Combining estimates (2.10), (2.13) and (2.14), we obtain where we used (2.12) and 0 < < 2µ 0 .6 V. T. Luong, N. T. Hue Thus the Gronwall-Bellman inequality yields the estimate where γ 0 = 2µ+ min{1;2µ 0 − } , ∈ (0, 2µ 0 ).Now multiplying both sides of this inequality by e −γt , γ > γ 0 , then integrating them with respect to t from h to ∞, we get By the Fubini theorem and γ > γ 0 , we conclude that where C is a constant.From the inequality (2.17), by standard weak convergence arguments, we can conclude that the sequence {u N } ∞ N=1 possesses a subsequence convergent to a function u h ∈ H1,1 (Q h , γ), which is a generalized solution of problem (2.2)-(2.4).
Proof of Theorem 2.1: Let k be a integer less than h, denote u k a generalized solution of the problem (2.2)-(2.4)with the replacement of h by k.We define u h in the cylinder Q k by setting Repeating this argument, we get This shows that {u h } is a Cauchy sequence in H1,1 (Q, γ).Hence, {u h } is convergent to u in H1,1 (Q, γ).Since u h is a generalized solution of problem (2.2)-(2.4),for any T > 0, we have that the equality holds for all v ∈ H1,1 (Q, −γ), v(x, t) = 0, t ≥ T. Using (2.18) when h → −∞, we obtain (1.5).
It means that u is a generalized solution of the problem (1.2)-(1.3).Using (2.6), we get From this inequality, sending h → −∞, we get (2.1).The proof of the theorem is completed.
Proof.It suffices to prove that the only solution of (1.2)-(1.3)with f ≡ 0 is u ≡ 0. To verify this, for any T > 0, fix 0 ≤ s ≤ T and set Then v ∈ H1,1 (Q, −γ) for any γ > 0. From the definition of generalized solution, we get where Hence, Using inequality (1.1) and the Cauchy inequality, we arrive at Let us write It follows readily from (2.19) that Choose T 1 so small that 1 − Ce 2γT 1 ≥ 1/2, then we have for all s ≤ T 1 .Consequently the Gronwall inequality implies u ≡ 0 on (−∞, T 1 ].In view of the uniqueness of the solution of the problem with initial condition (2.2)-(2.4),u ≡ 0 holds on R.
By the same arguments as in the proof of Theorem 2.1 together with inductive arguments (cf.[2]), we obtain the following theorem: Then for an arbitrary real number γ satisfying γ > γ 0 , the generalized solution u ∈ H1,1 (Q, γ) of problem (1.2)-(1.3)has derivatives with respect to t up to order h belonging to H1,1 (Q, γ), and where C is a constant independent of u and f .

Regularity of the generalized solution
We reduce the operator with coefficients at P ∈ l 0 , t ∈ R L (2) Clearly, the value ω(P, t) does not depend on the method by which L (2) 0 is reduced to its canonical form.The function ω(P, t) is infinitely differentiable and ω(P, t) > 0.Then, we have the following theorem.Theorem 3.1.Let the assumptions of Theorem 2.4 be satisfied for a given positive interger h + 1.Furthermore, assume α ∈ [0, 1] and 1 − α < π ω .Then the generalized solution u ∈ H1,1 (Q, γ) of problem (1.2)-( 1.3) has derivatives with respect to t up to order h, u t h ∈ H 2,0 α (Q, γ) and where C is a constant independent of u, f .
Proof.We will prove the assertion of the theorem by induction on h.Firstly, we consider the case h = 0.It is easy to see that u(•, t 0 ), t 0 ∈ R, is the generalized solution of the following problem: Multiplying both sides of the above inequality by e −t 0 γ , then integrating with respect to t 0 on R and using estimates from Theorem 2.4, we obtain Thus, the assertion of the theorem is valid for h = 0. Next, suppose that the assertion of the theorem is true for h − 1, we will prove that it also holds for k = h.Differentiating h times both sides of (1.2) with respect to t, we find By using the assumptions of the theorem and the inductive assumption, we obtain Multiplying both sides of (3.3) by e −t 0 γ , then integrating with respect to t 0 on R and using estimates from Theorem 2.4 again, we obtain It means that the assertion of the theorem is valid for k = h.The proof is completed.From now on, let the assumption of Theorem 2.4 be satisfied for a given positive integer h + 1.

Theorem 3.2. Assume that f t k
Then the generalized solution u of problem where C is a constant independent of u, f .
Proof.We have Therefore, u ∈ H 2 α (Q, γ) by Theorem 3.1 and Theorem 2.4.Moreover, we have Thus, the theorem is valid for h = 0. Suppose that the assertion of the theorem is true for h − 1, we will prove that it also holds for k = h.It is easy to see that Hence, we will prove that and By using Theorem 3.1, this holds for k = h.Suppose that it holds for k = h, h − 1, . . ., j + 1, we will prove that it holds for k = j.Returning one more time to (3.2) (h = j), we get

(by the valid inductive assumption for
, we obtain Multiplying both sides of (3.8) by e −γt , then integrating on R, we arrive at .

Asymptotics of the solution in a neighbourhood of the edge
According to the previous section, if k + 1 − α < π ω , α ∈ [0, 1] and f , f t , f tt ∈ H k α (Q, γ), then the solution u ∈ H 2+k α (Q, γ).Now we study the solution in the case π ω < k + 1 − α.In this case, we can find an asymptotic representation of u in the neighbourhood of l 0 : x 1 = x 2 = 0.In this section, we use the notations y 1 = x 1 , y 2 = x 2 , y = (y 1 , y 2 ), z i = x i+2 , z = (z 1 , . . ., z n−2 ), r = x 2 1 + x 2 2 and (r, ϕ) for the polar coordinates of the point y = (y 1 , y 2 ) To begin with, we present the following lemma.Lemma 4.1.Suppose that the following hypotheses are satisfied: Proof.Using (i), we get from Theorem 3.2 that u t s ∈ H k+1,0 On the other hand, we have where (2) where F ∈ H k α (Ω) for a.e.t ∈ R. Now we can apply Theorem 1' in [9] to get u(y, z, t) = c(z, t)r where It implies that this lemma holds for h = 0. Suppose the lemma is true for h − 1, we will prove that it also holds for k = h.Denoting v = u t h and differentiating both sides of (4.1) with respect to t, h times, we find The first term of the right hand side of (4.4) can be rewritten in the following form: From the assumptions of the lemma and the inductive assumptions, we find that F 1 ∈ H k α (Ω z ).Hence, from equality (4.4), we obtain where F 2 ∈ H k α (Ω z ).Employing the equality above, we get from (4.3) that

L
(2) where Therefore, where By the assumption (i), it implies that u is differentiable with respect to t.Then, it can be seen that the functions c(z, •), u 1 (y, z, •) are differentiable with respect to t. Combining (4.2) and (4.6), we conclude that The proof is completed.Now, we come to the main results.