ABOUT DECAY OF SOLUTION OF THE WAVE EQUATION WITH DISSIPATION

In this work, we consider the problem of existence of global solutions for a scalar wave equation with dissipation. We also study the asymptotic behaviour in time of the solutions. The method used here is based in nonlinear techniques.


Introduction
We will study the following evolution problem u tt − ∆u + a(x)u t = 0 in Ω × R + , (1.1) u = 0 on , ∂Ω × R + , (1.2) where Ω is an open bounded domain in R N with smooth boundary ∂Ω and a is a suitable, smooth and should not identically zero on Ω; besides, a can vanish in some part of Ω.
We define by the energy associated to the system (1.1)- (1.3).By Lemma 5.1 E is non increasing.Thus, we are interested in finding out what happens to E(t) as t goes to infinity and what is its rate of decay.
In this work, we study the existence of global solution and the asymptotic behaviour of the wave equation with dissipation, where the initial conditions satisfy the mth-order compatibility condition, with allows us to obtain a more regular solution.
We use the semigroup theory [17], [6] to prove the existence and uniqueness of solution to the problem (1.1)-(1.3),as well as its continuous dependency of initial data.Likewise, we study the regularity of this solution.
In section 4 we make a complete study of certain integral inequalities [15].Also we prove that and we use it in the proof of Lemma 4.2.Besides, in this section we introduce Lemma 4.4 which is an analogous version of Lemma 4.3.Making use of the multiplicative techniques [13], we obtain important estimations like (5.12), (5.16), (5.23).And by adapting the Conrad and Rao methods [1], we obtain the estimation (5.44) which allow us to prove Lemma 5.3 and then the hypothesis of the Lemma 4. 4.

Main Results
We state the result for existence of solution to the problem (1.1)- (1.3).
Theorem 2.1.Given (u 0 , u 1 ) ∈ (H 2 (Ω) ∩ H 1 0 (Ω)) × H 1 0 (Ω), there is only one solution u(x, t) of (1.1)- (1.3) Also, we will need the following result for regularity of the solution, for which we cite Kesavan [6] and Ikawa [5].We introduce the following definition Definition 2.1.The initial condition (u 0 , u 1 ) ∈ H m+1 × H m satisfies the mth -order compatibility condition associated to (1.1)-(1.3)if u k ∈ H m+1−k ∩ H 1 0 for k = 0, 1, . . ., m and u m+1 ∈ L 2 , (2. 1) where the sequence (u k ) k is defined by induction from (u 0 , u 1 ) by the formula Proposition 2.1.Let m ≥ 1 be an integer.Let us suppose that a ∈ C m−1 (Ω) and (u 0 , u 1 ) satisfies the mth-order compatibility condition associated to (1.1)- (1.3).Then, there is only one solution u(t) of the problem (1.1)- (1.3) and the linear application We observe that B is continuous and strictly increasing on [0, R  2 ] and that B(0 We will use the Lemma 4.4 in the proof the following Main Theorem. Theorem 2.2 (Main result).Let us suppose that a goes to zero at the boundary quickly enough so that there exist p > 0 and C > 0 such that Then, if (u 0 , u 1 ) satisfies the mth-order compatibility condition; there exists C > 0 which depends on the norm of the initial condition on where B −1 denotes the inverse function of B. Therefore, by the main theorem we obtain

Remarks of Theorem
r k with k > 0 then, we can apply remark 2.1.Therefore, by the main theorem we get

Existence of solution
From the equation (1.1) writing v = u t we get: Theorem 3.1.The Operator A defined above generates a contraction semigroup S(t) on the Hilbert Space H.
Proof.-Observe that D(A) is dense in H.We will prove that where Im(z) is the imaginary part of z ∈ C. Taking the real part of the equality (3.2), we have Now, we will prove that 0 ∈ ρ(A).In fact, let F = (f, g) T ∈ H 1 0 (Ω) × L 2 (Ω) = H, we will prove that there is U = (u, v) T ∈ D(A), such that AU = F .Let us consider the equations By standard results on Elliptic linear equations, we have that (3.5) has only one solution u ∈ H 2 (Ω) ∩ H 1 0 (Ω).From (3.3) we obtain v = f .That is, A is an onto map.
We claim that A is one to one.In fact, let AU = 0 then Replacing (3.6) in (3.7) we have ∆u = 0 and using the Green's Identity we have |u| 2 hence u = 0 in H 1 0 (Ω).From (3.6) we have that v = 0. Therefore U = 0. i.e.A is one to one.
Thus, there is A −1 : H −→ D(A) because A is one to one and H is the image of A. Now, we will prove that A −1 is bounded.Multiplying the equation (3.5) by u and integrating on Ω, we have R Ω ∆uudx, using the Holder and Poincaré inequalities, we obtain Then, taking > 0 such that 1 − C p > 0 we have Hence we have Thus, using (3.8), v = f , and the Holder and Poincaré inequalities we get which allow us to say that A −1 is bounded.Now, by the Lummer-Phillips theorem, we have that A is the infinitesimal generator of a C 0 semigroup of contraction on H : S(t).
From these two remarks, we get the following result.
Proposition 3.1.There exists only one solution of (3.1), Now, we will finish the proof of Theorem 2.1 Since U 0 = (u 0 , u 1 ) ∈ D(A), by Proposition 3.1 we obtain that there exists Since U satisfies (3.1) we have u t = v and v t = ∆u − av.By one hand, we have Since A is closed and there exists A −1 (It was proved in 0 ∈ ρ(A)), then A −1 is also closed.

Integral Inequalities
Lemma 4.1.Let E : R + → R + be a no increasing function and φ : R + → R + be an strictly increasing C 1 function such that Let us suppose that there are σ ≥ 0, and w > 0 such that ∀s ≥ 0 , Then, E satisfies the following estimates: (t) , for all t ≥ 0. (4.3) Now, we prove for E(0) = 1.
We introduce the following function f : then f is no increasing.Making a change of variable and using (4.2) with E(0) = 1, we obtain the following: Since lim T →∞ φ(T ) = +∞, then f satisfies: 1+σ) dτ .So, h is well defined, no increasing, no negative and satisfies the following differential inequality.
By other hand, from (4.5) and f (0 In fact, in the last inequality of (4.8) it is used (4.7).
On the other hand, since f is no increasing, we have that f (t).≤ Since E(t) = f (φ(t)), by using (4.9), we get (4.3).
In analogy to this Lemma, we have the next version.Let us assume that there are σ > 0, σ 0 > 0 and c > 0 such that ∀s ≥ 1 , Then there exists C > 0 depending continuously on E(1) satisfying

Using the multiplier method
Let (u 0 , u 1 ) ∈ H m+1 (Ω) × H m (Ω) satisfying the mth -order compatibility condition.Then, the regularity given by (2.3) justifies the calculus we are going to do.
Proof.-Multiplying the equation (1.1) by u t and integrating on Ω, and using the Green Identity, we have then the result holds.Let σ ≥ 0, and φ : R + → R + be a concave and increasing C 2 function.Let w be a neighborhood of the boundary ∂Ω.Lemma 5.2.Let h : Ω −→ IR N be a C 1 vector field, σ ≥ 0 and 0 ≤ S ≤ T < +∞.Then we have, (5.2) Proof.-Multiplying the equation (1.1) by E σ φ 0 2h • ∇u and integrating on [S, T ] × Ω we have R Ω 2h(∇u)u tt dx and integrating by parts we obtain Using the Green Identity we have Replacing I 1 and I 2 on the equality (5.3) we obtain Using the fact that h∇(u 2 t ) = div(hu 2 t ) − (div h)u 2 t and the Divergence Theorem )dx, we have By other hand, we have And since div(h|∇u| 2 ) = (divh)|∇u| 2 +h∇(|∇u| 2 ) and using the Divergence Theorem, we have Using (5.5) and (5.6) we obtain Replacing I 3 and I 4 in the equality (5.4), we will have the result.

holds.
Proof.-Let K 1 be a compact of Ω such that Ω − K 1 be a compact set on w.
Define h(x) := β(x)m(x), where m(x) = x − x 0 and β is a C ∞ function whoose support is compactly in Ω and equal to 1 on K 1 .Since φ 0 is no increasing and positive then φ 0 is bounded on IR + (i.e.|φ 0 (t)| ≤ M ).Now, we apply (5.2) to this h and get (5.8) By other hand, using that ≤ c 0 E(S) σ+1 .(5.9) And, since E(T ) < E(S) and by Holder Here we need to make the following estimate where will be considered little enough.Using the inequalities (5.9), (5.10) and (5.11) on (5.8) we have that there exists C > 0 such that with little enough.
Integrating by parts the expression: we have By the Poincaré inequality we have Using this in (5.9) we obtain With a similar proof to (5.10), we also obtain Adding (5.12) and (5.13), taking < 1 and using (5.14) and (5.15) we have We want to eliminate the last term of (5.16).To do this, we construct a function ξ ∈ C ∞ (Ω) such that ξ = 1 in Ω − K 1 and ξ = 0 outside w.
We multiply the equation (1.1) by ξu and integrate it on Ω; then we multiply this expression by E σ φ 0 , and integrate on [S, T ], and integrating by parts we get Using the Green Identity and ∇(u 2 ) = (∇u) u + u∇u we have Replacing (5.18) on (5.17) we obtain from where Since ξ is bounded, in a similar way to (5.9), we obtain (5.21) Also, using the fact of ξ is bounded, similarly to (5.10) we get And, using (5.21) and (5.22) in (5.20) we obtain (5.23) Now, to eliminate the last term of ( 5), we will adapt the Conrad and Rao [1] method.
We start with an arbitrary function β ∈ C ∞ (R N ) be such that 0 ≤ β ≤ 1, β = 1 on w and β = 0 outside a neighborhood of w (see [4], Theorem 1.2.2, or [2] p.p 3489).Now, fix t and consider the solution z to the elliptic problem: Multiplying the equation (5.24) by z, integrating on Ω and using the Green Identity, we have: hence, using the Holder and Poincaré inequalities, we obtain (5.27) Similarly to (5.26) we have Differentiating with respect to t to the equation (5.24) we have the problem ∆z t = β(x)u t in Ω (5.30) (5.31) Multiplying (5.30) by z t , integrating on Ω and using the Green Identity we obtain hence, using the Holder and Poincaré inequalities, we obtain Using the Green Identity and the fact that z is solution of (5.24)-(5.25)we obtain (5.37) Using (5.37) on (5.36) we have We can observe that the following inequality holds (5.39) By other hand, let η > 0, using (5.35) we obtain where η will be taken little enough.Also, we have where γ will be taken very little. Since (5.42) Using (5.39), (5.40) and (5.42) in (5.38), we have Taking γ < 1, from (5.43) we have that there exists C > 0 such that ∀η > 0 holds, where η is little enough.

Finishing the proof of the Theorem
Let ρ : t −→ ρ(t) a decreasing function which goes to zero as t goes to the infinite.Later on, we choose ρ.
Let us define the function α by α(r, t) Let p > 0 such that (2.6) holds.Then, using the Jensen Inequality and (6.4) we will estimate the last term of (5.7).Now, choosing ρ and φ carefully, we will estimate the last term of (6.9).The Choice of the function ρ.

2 ]
where D k denotes any kth order partial differentiation with respect to t and x.Let us suppose that Ω is the open ball in R N centered on 0 and the radius R. Let us also assume that ∀ |x| ≥ R 2 , a(x) := ã(|x|), where ã:[ R 2 , R] → R + is a strictly decreasing function which satisfies ã(R) = 0. (Note that R 2 could be replaced by any R − with > 0).Set ∀r ∈ [0, R , b(r) := ã(R − r) and B(r) := rb(r).(2.5)