EXISTENCE AND UNIQUENESS OF SOLUTIONS FOR DAMPED LINEAR HYPERBOLIC EQUATIONS WITH DIRICHLET BOUNDARY CONDITIONS

We consider damped linear hyperbolic equations with Dirichlet boundary conditions. We prove the existence, uniqueness, and regularity of the solution. We apply semi-discretization in time technique. AMS Subject Classification: 35A01, 35A02, 35L20


Introduction
In this paper, we study the initial boundary value problem for the following damped linear hyperbolic equations with Dirichlet boundary conditions.
x) = 0, on Σ, u(0, x) = u 0 (x), ∂u(0,x) ∂t = u 1 (x), x ∈ Ω. (1) where η is the constant damping coefficient, η ∈ R and T finite, Ω is a bounded open domain in R n , n ≥ 1, with a smooth boundary ∂Ω.We denote by Q the cylinder of R n x × R t , Q = Ω×]0, T [ and by Σ the lateral boundary of Q, Σ = ∂Ω×]0, T [, ∆u = Σ n i=1 ∂ 2 u ∂x 2 i is the Laplacian, and linear function f : Ω×]0, T [−→ R for i = 0, 1, the functions u i : Ω −→ R, are given.we find a function u = u(x, t), is a real-valued satisfies (1).This Problem has its origin in a physical problem, we study a model that describes the transverse vibrations of a membrane Ω fixed at its ends and in the presence of damping η.Let u(x, t) be the vertical position of x ∈ Ω at time t ∈ [0, T ], is retarded by a damping force proportional to the velocity of the membrane, then u satisfies (1).
This problem has been already investigated by many authors.For example in ( [2]), ( [7]), ( [17]).We define some function spaces required to establish the existence and uniqueness of solution to (1).We use the function spaces for any 1 ≤ p < ∞, L p (Ω) is the space of real measurable functions u : Ω −→ R for the Lebesgue measure dx, it is a Banach space for the following norm if X is a Banach space, 1 ≤ p < ∞, L p (0, T ; X) is the space of measurable functions u of ]0, T [ into X for the Lebesgue measure dt, which is Banach space for the following norm L ∞ (0, T ; X) is the space of measurable functions from ]0, T [ into X which are essentially bounded, the space is Banach for the following norm u L ∞ (0,T ;X) = sup t∈]0,T [ ess u(t) X we denote by C ([0, T ]; X) the space of continuous functions from [0, T ] into X, the space is Banach for the following norm u C ([0,T ];X) = sup t∈[0,T ] u(t) X , and by C k ([0, T ]; X), k ∈ N the space of k times continuously differentiable functions from [0, T ] into X, it is a Banach spaces for the following norm we denote by C ∞ (Ω) the space of infinitely times continuously differentiable functions on Ω.The space C ∞ (Ω) of real functions on Ω, with a compact support in Ω, is denoted by D(Ω), as in the theory of distributions of L.Schwartz in ( [8]), D ′ (Ω) is the space of distributions on Ω.
We introduce the Sobolev spaces, for m ∈ N, 1 ≤ p ≤ ∞, W m,p (Ω) is the space of functions u in L p (Ω) whose distribution derivatives of order ≤ m are in L p (Ω).This is a Banach space for the norm where When p = 2 we write W m,2 (Ω) = H m (Ω) and this is a Hilbert space for the scalar product (u, v) we use the space H 1 0 (Ω) for study the problem (1) we introduce the space V is a Hilbert space for the scalar product and that the corresponding norm with dual V ′ = H −1 (Ω), the corresponding norm on V ′ is for any l ∈ V ′ and u = 0, where ., . is the scalar product between V and V ′ .
To given linear continuous operator A ∈ L (V, V ′ ), we can associate a bilinear continuous form a on V by setting such that L (V, V ′ ) is the space of linear continuous operators from V into V ′ .conversely, to given a bilinear continuous form a on V , we can associate with a a linear continuous operator A from V into V ′ , and from the properties of a that A is linear continuous, and by the continuity of a if then The Riesz representation theorem to show that each a linear continuous form on H can be represented with the aid of scalar product.Let H ′ the dual space of H, to given φ ∈ H ′ there exists a unique f ∈ H such that the application φ −→ f is an isomorphism and isometric that allows us to identify H to the dual space H ′ .In general, but not always, it is also convenient to identify H to its dual H ′ .We write position typical where is not place of performance this identification.Let H = L 2 (Ω), and V = H 1 0 (Ω) is a dense in L 2 (Ω) since that V it is a Banach space reflexive, we assume that the canonical injection of V in H being continuous, then identify, H ′ ≡ H and H ⊂ V ′ from the following assertion: For where each space is dense in the following, the injections being continuous.We recall some basic results for using in the proof of our main results.

Preliminaries
Proof.we have the inequality by backward we obtain : pass to the limit as n −→ +∞, we obtain Proof.We find u solution of the problem (1) in the space L ∞ (0, T ; X) then we need the derivative ∂u ∂t in the space L ∞ (0, T ; X) we prove ∂u ∂t ∈ L p (0, T ; X) if u ∈ L p (0, T ; X) where 1 ≤ p ≤ ∞.
if u ∈ D ′ (0, T ; X), the distributional derivative is defined by if u ∈ L p (0, T ; X), the corresponding distribution is also defined by u from ]0, T [ into X, such that the integral u(ϕ) ∈ X; we can also defined ∂u ∂t ∈ D ′ (0, T ; X) by (9).
Let V is a Banach space separable and reflexive and K is a closed convex set in V .Theorem 3. We assume that K is a closed convex set unbounded in V .Let A is a pseudo-monotone operator from K into V ′ , and coercive in the following sense: then, for f ∈ V ′ , there exists u ∈ K such that Proof.We note that A is pseudo -monotone from V into V ′ if satisfies the following conditions: First: A is bounded, Second: as j −→ +∞, u j tending to u weakly in V and lim sup A(u j ), We give the following Theorem Theorem 4. We assume that K is a convex closed bounded noempty.Let A is a operator pseudo-monotone from K into V ′ .Then for f ∈ V ′ , there exists u in K such that The proof of the Theorem is in ( [3] since K R is a closed convex and bounded, then from Theorem 4, there exists choosing (11) we deduce from (10), that since K is weakly closed, u ∈ K.We have and from the pseudo-monotone, and since we deduce from (12) that

Existence and Uniqueness of Solutions
Theorem 5. Assume that Ω be a bounded open.We give f, u 0 , u 1 with There exists a unique solution u satisfies Proof.Proof of the existence.The proof of this Theorem will be made in three steps.
Step 1: The existence is proved by the Faedo -Galerkin method, in ([3]), we take u ′ = ∂u ∂t , u ′′ = ∂ 2 u ∂t 2 , since the space V = H 1 0 (Ω) is a separable space.We introduce a sequence of functions where g im , are obtained by the conditions (u ′′ m (t), w j ) + η(u ′ m (t), w j ) + a(u m (t), w j ) = (f (t), w j ), where the system (22) of linear ordinary differential equations is given with the initial conditions, as m −→ +∞ From the linearly independent of w 1 , • • • , w m , we have det(w i , w j ) = 0, i = 1, • • • , m and j = 1, • • • , m then from the general results on the systems of differential equations, these results guarantees the existence of a solution of (22) − (24) − (25) in the intervalle [0, t].The following a priori estimates show that t = T .
Step 2: We multiply (22) by g ′ jm , add these relations for then after an integration on t and using Cauchy-Schwarz inequality, we obtain where C > 0 is independent of m from (13) we have we use the Lemma 1, we obtain if η < 0 we conclude from (30) that we use the Lemma 1, we obtain then we conclude that t = T , from (32) − (33) we obtain the result, letting m −→ +∞, Step 3: Pass to the limit From Dunford-pettis theorem in ( [16]) to show that the space L ∞ (0, T ; H 1 0 (Ω) be a given with dual L 1 (0, T ; H −1 (Ω)) and the space L ∞ (0, T ; L 2 (Ω)) be a given with dual L 1 (0, T ; L 2 (Ω)) by a consequence there exists a subsequence u µ of u m such that u µ −→ u weakly star in L ∞ (0, T ; and almost everywhere pass to the limit in (22) and using for m = µ, let j is a fixed and µ > j, then from (22 and we conclude from (37) that this for j is a fixed arbitrary.We multiply (38) by g jm , add these relation for then u satisfies ( 18) and ( 17) − ( 16).For show that ( 19) is satisfying, from (35) − (36) and the Lemma 2 we have, u µ (0) −→ u(0) weakly in L 2 (Ω), and from (24 Proof.Suppose that on a measurable set E, we note that Let N is an increasing sequence tending to +∞, we introduce indeed, we have u µ and u are functions in From ( 40), and from ( 25), for any j, then we have (20).
Remark 9.In the case of the uniqueness the sequence u m of approximate solutions converges to u.

A Regularity Result
Theorem 10.The hypotheses are those of theorem 5 with another then there exists a unique solution u of (18) Proof.Proof of the Existence The Proof of this Theorem will be made in two steps, the existence is proved by the Faedo-Galerkin method, in ( [3]).From the approximate solution u m of ( 22) − (24) − (25), we take w j , j = 1, 2, • • • , m is a basis in the space H 1 0 (Ω) H 2 (Ω), from (24) − (25) we assume that We prove in step 1 an additional a priori estimate to show that the existence of a solution with (55) − (56), and we prove (54) in step 2 by using the equation (18).
For justify the previous conclusion the uniqueness applying the same argument in uniqueness Theorem 8, then the sequence u m of approximate solutions converges to u.

Semi Discretization and Variational inequalities
In ( [3], P. 432), we apply semi-discretization in time, to establish the existence and uniqueness of solution, then we give the following Theorem.Such that from (8) we have and from (7), we have Theorem 12.We assume that (77) − (78) are satisfied.Let K is a closed convex set in V .We give There exists a unique solution u, such that And in the cases 1.,2.We obtain the inequality And in the cases 2., 3. We obtain the inequality And in the cases 2.,4.,5.We obtain the inequality Proof.Semi-discretization in time We introduce N is an integer fixed in N and u n an approximation to u at time nk.We introduce we take u 0 = u 0 (89) and we define u n by (90) The system (90) is an elliptic variational inequality, has a unique solution.Indeed, (90) is equivalent to, for any then by applying the Theorem 3, to the operator Where I is the identity operator, then we prove that operator ∂x i dx, v 0 ∈ K, and from ( 4) − (7) we obtain that and we prove that operator −∆ − ( 2 k 2 + η k )I is pseudo monotone, first we prove that operator is bounded then for any u, v ∈ K, from (5) we have then by applying the Theorem 3, we deduce the system (90) has a unique solution.We say that the system (90) is a semi-discrete approximation of the inequalitys (84), ( 85), (86).

Proof of the Existence
We introduce then we prove the Lemma Lemma 13.We take k −→ 0, we have Proof.We refer to ( [13]), ([3], P. 222), ( [5]) for the proof of Lemma 13. we want to solve the initial-value problem such that η ∈ R Where f ∈ L 2 (0, T ; H), u 0 ∈ K, u 1 ∈ H. Since the space V is a separable space and consider a sequence of linearly Using the Faedo-Galerkin method we define for each k an approximate solution u k of (93) − (94) then Where u 0k is the projection in V of u 0 on the space spanned by w 1 , • • • , w k , and u 1k is the projection in V of u 1 on the space spanned by w 1 , • • • , w k .Equations ( 95)−( 96)−( 97)−( 98) are equivalent to a linear initial-value problem for an ordinary k-dimensional differential equation.They possess a unique solution defined for all time and in particular on [0, T ], the function A priori estimates are obtained by multiplying (96) by g ′ jk and summing these relations for j = 1, • • • , k.We obtain We use the Lemma 1, we obtain

Then we have (92).
Thus there exists a subsequence, remain denoted u k , and u, such that 0, T ; H) weak-star .Then passing to the limit in (95) − (96) − (97) − (98) we see that u is a solution of (93) − (94) which satisfies (101).To conclude the proof of existence it remains to show the continuity properties u ∈ C ([0, T ]; V ), u ′ ∈ C ([0, T ]; H).We give the following Lemma Lemma 14.Let X and Y be two Banach spaces such that with a continuous injection.If a function ϕ belongings to L ∞ (0, T ; X) and is weakly continuous with values in Y , then ϕ is weakly continuous with values in X.
The proof of the Lemma 14 is in ( [12]), and in ( [14], Lemma 1.4, Chapter III).It follows from the Lemma 14 and (101) that u is weakly continuous from [0, T ] in V .Similarly, we infer from (93) that We give the following Lemma Lemma 15.Let X be a given Banach space with dual X ′ and let u and g be two functions belonging to L 1 (0, T ; X).Then the following three conditions are equivalent (i) u is almost everywhere equal to a primitive function g, there exists ξ ∈ X such that in the scalar distribution sense on ]0, T [.
are satisfied we say that g is the X-valued distribution derivative of u, and u is almost everywhere equal to a continuous function from [0, T ] into X.
The proof of the Lemma 15 is in ( [14], Lemma 1.1, Chapter III).From Lemma 15, then shows that u is continuous from [0, T ] in V ′ , Lemma 14 and (101) imply that u ′ is weakly continuous from [0, T ] in H.We give the following Lemma Lemma 16.We assume that w is such that Then, after modification on a set of measure zero, u is continuous from The proof of the Lemma 16 is in ( [13], P.79).We deduce from Lemma 16 that u satisfies an equation similar to (100), namely This shows that the function is continuous on [0, T ].In conjunction with the above properties of weak continuity, we conclude that u ∈ C ([0, T ]; V ) and u ′ ∈ C ([0, T ]; H), then we have (82) For show that (83) we have if the set K is a closed and convex of functions v ∈ C ([0, T ]; V ) such that v(t) ∈ K almost everywhere, then we have u k ∈ K for any k and since K is weakly closed in C ([0, T ]; V ), we have u ∈ K.
For show that (84), ( 85), (86), we consider the function v satisfies we put We note that We define we have We take v = v n in the system (90), and we multiply by k 2 we obtain that From the properties of the norm in L 2 (Ω), we have the inequality if we have the case 1.: 2 < 0 by using the case 1., on the member of the equality (115) and using the inequality (116), we obtain the inequality if we have the case 2.: > 0 by using the case 2., on the member of the equality (115) and using the inequality (116), we obtain the inequality by using the inequalitys (114) − (117) − (118), on the equality (115) we obtain summing to n, we deduce with ( 110) − (111) − (113), we conclude of (120) the inequality for any v satisfies (109).if v is given as in the inequality (84), there exists v j satisfies the conditions (109) and such that ).We take v = v j in (122) and pass to the limit, we conclude that (84).If we have the case 3.: < 0 by using the case 3., on the member of the equality (115) and using the inequality (116), we obtain the inequality by using the inequalitys (114) − (123) − (118) on the equality (115) we obtain summing to n, we deduce with (110) − (111) − (113), we conclude of (125) the inequality for any v satisfies (109).If v is given as in the inequality (85) there exists v j satisfies the conditions (109) and such that ).We take v = v j in (127) and pass to the limit, we conclude that (85).If we have the case 4.: > 0 and the case 5.: > 0 in the cases 4., 5., we use the inequalitys (114) − (118) on the equality (115) we obtain the inequality summing to n, we deduce with (110) − (111) − (113), we conclude of (129) the inequality and for any v satisfies (109).If v is given as in the inequality (86) there exists v j satisfies the conditions (109) and such that ).We take v = v j in (131) and pass to the limit, we conclude that (86).

Proof of the Uniqueness Regularity parabolic and variational inequalities hyperbolic
We approach parabolic equations by elliptic equations the following step is to approach hyperbolic equations by parabolic equations, this is the regularity parabolic method, that allows us to proved the Uniqueness.We apply this method to the evolution inequalities of type hyperbolic or related to operators well-posed of a sense of Petrowski.Hypotheses Let V , H are Hilbert spaces with The scalar product on V is given by (138), the scalar product in H is given by then there exists α 0 > 0 such that The semigroups G(s) and g(s) We give the semigroup G(s) in L 2 (V ), L 2 (H) and L 2 (V ′ ).G(s) is a semigroup of contractions in L 2 (H).We denote by −Λ is the infinitesimal generator of G(s), and D(Λ; L 2 (H)) is the domain of Λ in L 2 (H).We can associate to G(s) the semigroup g(s) in V , H , V ′ , is given by and we denote by −L is the infinitesimal generator of g(s), is given by with the domain Remark 19.The variational inequality (154) is not of type parabolic because A is restrict to V is not coercive on V , but is only coercive on H , this is the situation typical of hyperbolic operator or well-posed of a sense of Petrowski.
Remark 20.Before we give applications expounding (154), since K = K 1 × K 2 , then from (154) we have the inequalitys In particular, K 1 = L 2 (V ) then from (155) we deduce the equation Λu Remark 21.Third a priori estimate in the proof of the Theorem 18 is not satisfying if then (148) is not utility.
Remark 22. Since A is given by ( 140), but with k = 0 we have and perform the following hypothese : and if u j ∈ D(Λ; L 2 (V )) and (Λ + ǫ)v j is remain in bounded of L 2 (V ), then v j is remain in a bounded set of L 2 (V ), for any j, as j −→ 0. In these conditions, the Theorem 18 is also satisfying.Indeed, by taking then there exists u ǫ in K 1 × K 2 and D(L; V ) such that there exists and then But u ǫ2 ∈ K 2 bounded in L 2 (V ) and then u ǫ1 is remain in a bounded set of L 2 (V ) and bounded of D(Λ, L 2 (V )).
Then we have the results of first and third a priori estimate in the proof of the Theorem 18, second a priori estimate is unchanged.
Remark 23.From the proof of Theorem 18 can be replaced The hypotheses (150) − (151) by the following hypotheses respectively, there exists β ∈ R, such that ∀s ≥ 0, e βs G(s there exists ρ > 0, such that, ∀s ≥ 0, ∀v ∈ K First application From the Theorem 18 we prove as application the following Theorem.
Proof.We apply the Theorem 18 in the following conditions Here the conditions of the Theorem 18 are satisfied, choosing k > 0. The operator Λ is Λ = ∂ ∂t , with domain the null functions for t = 0, here by using (157), then we have the existence and uniqueness of the couple u 1 , u 2 with u 1 ∈ D(Λ; L 2 (V )), such that, As K 2 is cone of summit the origin, (172) is equivalent to Using the definition of K 2 , we conclude that If we multiplying by v 2 and integration by parts, we deduce of (174) that Then put w i = e kt u i , i = 1, 2.

Second application
We consider the problem with periodic solutions in t of inequality of type (166)− (168).We take that the change w i = e kt u i , i = 1, 2, exterminate the periodicity in t then we will verify this.We give the following Theorem as application on the Theorem 18 and the Theorem 24 Theorem 25.We give the function f = f (x, t), with Let k = η > 0.
There exists a unique function u, satisfies (165) and Proof.Applying the Theorem 18 with the same hypotheses of the Theorem 24, but the semigroup G(s) is given by G(s)ϕ(t) = ϕ(t − s + T ), if t ≤ s, ϕ(t − s), if t ≥ s.
We obtain the existence and uniqueness of the couple {u 1 , u 2 } with and (171) − (172).Then interpret (172) as in the Theorem 24 and that u = u 1 satisfying the conditions of the Theorem 24, as we have equivalent of the research of u, of {u 1 , u 2 } we have further the uniqueness.
Then we obtain the Uniqueness in the Theorem 12 when η > 0.
To illustrate our results we consider the following example.Then, we have the following approximation v 1 j for u(x j , △t) To write (185) in a more compact form, we let v m ∈ R n be the vector where I ∈ R n,n , is the identity matrix, and

w ′ 1 − w 2 = 0, w ′ 2 −Then u = w 1
∆w 1 = e kt f = f * , w 2 ≥ 0, on Σ, satisfying the conditions of Theorem 24 , with f is replaced by f * .As we have equivalent of the research of u, of {w 1 , w 2 } and {u 1 , u 2 } we have further the uniqueness.