Tamtam Proceedings - lamsin

Goursat boundary value problem for hyperbolic equation 127-The regularizing operators A −1ε (t) are strongly differentiable for almost t, in H,commute with A(t), ∥ A−1ε (t) ∥ L(H)≤ 1 and when ε tends to 0 we have for all u ∈ H:∣ εA(t)A−1ε (t)u ∣ ∣ = ∣u − A−1ε (t)u ∣ → 0. (P 1) (4)-The operators A(t)A −1ε (t) are strongly differentiable for every t ∈ D and∂(A(t)A −1ε∂t i(t))= −1ε∂A −1ε∂t i(t)=−A(t)A −1εWe integrate by parts the double real part of the expressione c(s1+s2−t1−t2) × ( Lu, A −1ε (t)u )(t) ∂A−1 (t)A(t)A −1ε (t), i = 1, 2. (P 2)∂t i(5)over the domain D s = ]0, s 1 [ × ]0, s 2 [ ⊂ D, then we use δ−Cauchy inequality and weapply the properties of regularizing operators when ε goes to 0. Finally the inequality (3)follows by passing to the supremum over (s 1 , s 2 ) ∈ D and the constant C is equal toe T1+T2 .Lemma 2. The following statement is true in the view of Theorem 1: The operator L hasa cloture L with domain of definition D(L) = D(L).Proof. Since the operators l 1 u and l 2 u are continuous it suffices to prove that foru n ∈ D(L), u n → 0; Lu n → f ∈ L 2 (D, H) ⇒ f = 0. (6)From the density of D(L) in E (see Lemma 1) it follows that f = 0.So a function u is in D( L) _ if there exist a sequences (u n ) ∈ D(L) and an element_F ∈ F such that ‖u − u n ‖ 1→ 0 and ‖Lu n − F ‖ 2→ 0 i.e: Lu = lim n→∞ Lu n thenD(L) = D(L).Definition 1. The solution of equationLu = F,is called a strong generalized solution of the considered problem.By passing to the limit we extend the inequality (3) to strong solutions u ∈ D ( L ) :‖u‖ 2 1 ≤ C ∥ ∥ Lu 2, ∀u ∈ D(L). (7)2Remark 2. From the inequality (5), we deduce that the strong generalized solution of theGBVP, when it exists is unique, depends continuously on the data (f, ϕ), the range R(L)of the operator L is closed in F, R(L) = R(L) and (L) −1 = L −1 .To prove the existence of the strong generalized solution, it remains to prove that therange R (L) is dense in the Hilbert space F, which is equivalent to R (L) ⊥ = {0} .TAMTAM –Tunis– 2005