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128 A. Guezane-Lakoud et al.5. Existence of the strong generalized solutionTheorem 2. Assume that the conditions of the Theorem 1 hold. Then for any u ∈D( __ L) and (f, ϕ, ψ) ∈ F, there exists one and only one strong generalized solution u =L −1 _(f, ϕ, ψ) to the GBVP (1)-(2) satisfying‖u‖ 2 1 ≤ C ‖(f, ϕ, ψ)‖2 2, ∀u ∈ D(L).Proof. Let V = (v, ϕ, ψ) ∈ R(L) ⊥ and F = (f, ϕ, ψ) = Lu = (Lu, l 1 u, l 2 u), u ∈D(L), then from(F, V ) = 0 ⇒ (Lu, v) L2(D,H) + (ϕ, l 1 u) L2 (]0,T 1 [,H) + (ψ, l 2u) L2 (]0,T 2 [,H) = 0,we shall prove that V = 0.First step. Let u ∈ D(L) = {u ∈ D(L) : l 1 u = l 2 u = 0}, then(F, V ) = 0 ⇒ (Lu, v) L2(D,H) = 0. (8)We set u = A −1ε (t)h, where h is an arbitrary element of L 2 (D, H) in (6), take thedouble real part, using the properties of the regularizing operators A −1ε (t) when ε tendsto 0 and putting h = v, we obtain:∫ T20|v| 2 t 1=T 1dt 2 +∫ T10∫ ∫|v| 2 t 2=T 2dt 1 + 2consequently from condition (a) we have v = 0.Second step. Let u ∈ D(L), thenD∣∣∣A 1 2 ∣∣2(t)v dt1 dt 2 = 0,(F, V ) = 0 ⇒ (ϕ, l 1 u) L2 (]0,T 1 [,H) + (ψ, l 2u) L2 (]0,T 2 [,H) = 0.Since the range of the operator (l 1 , l 2 ) is everywhere dense inL 2 (]0, T 1 [ , H)×L 2 (]0, T 2 [ , H), then ϕ = ψ = 0, so, V = 0. This achieves the proof ofthe Theorem 2.6. References[1] N. I. BRICH & N. I. YURCHUK, Goursat’s problem for abstract second order linear differentialequations. Diff. Uravn. 7. 1017-1030. (1971).[2] A. GUEZANE-LAKOUD, On a class of hyperbolic equation with abstract non-local boundaryconditions. Internat. J. Appl. Sci. Comput.Vol 8, N ◦ 2, 110-118, (2001).[3] A. GUEZANE-LAKOUD, Abstract variable domain hyperbolic differential equations.Demonstratio Mathematica. N 4. V 37. 884-892, (2004).[4] A. GUEZANE-LAKOUD, On Goursat boundary value problem. 6TH Pan-African Congressof Mathematicians. September.01-06, 2004, Tunis , Tunisia.TAMTAM –Tunis– 2005