EXISTENCE FOR A CLASS OF DISCRETE HYPERBOLIC PROBLEMS RODICA LUCA Received 1 November 2005; Revised 10 March 2006; Accepted 7 April 2006 We investigate the existence and uniqueness of solutions to a class of discrete hyperb olic systems with some nonlinear extreme conditions and initial data, in a real Hilbert space. Copyright © 2006 Rodica Luca. This is an open access article distributed under the Cre- ative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction Let H be a real Hilbert space with the scalar product ·,· and the associated norm ·. In this paper we will investigate the discrete hyperbolic system du j dt (t)+ v j (t) − v j−1 (t) h j + A u j (t) f j (t), dv j dt (t)+ u j+1 (t) − u j (t) h j + B v j (t) g j (t), j = 1,N, t ∈ [0,T], in H,(S) with the extreme conditions v 0 (t) =−α u 1 (t) , u N+1 (t) = β v N (t) , t ∈ [0,T], (EC) and the initial data u j (0) = u j0 , v j (0) = v j0 , j = 1,N,(ID) where N ∈ N, h j > 0, j = 1,N,andα, β, A, B are operators in H, which satisfy some assumptions. Hindawi Publishing Corporation Advances in Difference Equations Volume 2006, Article ID 89260, Pages 1–14 DOI 10.1155/ADE/2006/89260 2 Existence for a class of discrete hyperbolic problems This problem is a discrete version with respect to x (with H = R )oftheproblem ∂u ∂t (t,x)+ ∂v ∂x (t,x)+A u(t,x) f (t,x), ∂v ∂t (t,x)+ ∂u ∂x (t,x)+B v(t,x) g(t,x), 0 <x<1, t>0, in R,(S) with the boundary conditions v(t,0) =−α u(t,0) , u(t,1) = β v(t,1) , t>0, (BC) and the initial data u(0,x) = u 0 (x), v(0, x) = v 0 (x), 0 <x<1. (IC) The above problem has applications in electrotechnics (the propagation phenomena in electrical networks) and mechanics (the variable flow of a fluid)—see [7, 8, 13]. The system (S) subject to various boundary conditions has been studied by many authors: Barbu, Iftimie, Moros¸anu, and Luca, in [4, 5, 9, 11, 12]. Using an idea from [14] we dis- cretize the problem (S)+(BC)+(IC) in this way: let N be a given integer (N ≥ 1) and h = 1/(N + 1). In a first stage we approximate the system (S) and the boundary condi- tions (BC)by ∂u ∂t (t,x)+ v(t,x) − v(t,x − h) h + A u(t,x) f (t,x), x ∈ h,(N +1)h , ∂v ∂t (t,x)+ u(t,x +h) − u(t,x) h + B v(t,x) g(t,x), x ∈ (0,Nh), t>0, v(t,0) =−α u(t,0) , u(t,Nh) = β v(t,Nh) . (1.1) We look for u and v of the form u(t,x) = N j =0 u j (t)ϕ j (x)andv(t,x) = N j =0 v j (t)ϕ j (x), where ϕ j (x) = χ [jh,(j+1)h) (x) = ⎧ ⎨ ⎩ 1, x ∈ jh,(j +1)h , 0, x ∈ jh,(j +1)h . (1.2) We write f , g, u 0 , v 0 as f (t,x) = N j=0 f j (t)ϕ j (x), g(t,x) = N j=0 g j (t)ϕ j (x), u 0 (x) = N j=0 u j0 ϕ j (x), v 0 (x) = N j=0 v j0 ϕ j (x), (1.3) where f j (t) = f (t, jh), g j (t) = g(t, jh), u j0 = u 0 (jh), and v j0 = v 0 (jh). Rodica Luca 3 Then for u j and v j we obtain the system u j + v j − v j−1 h + A u j f j , j = 1,N, v j + u j+1 − u j h + B v j g j , j = 0,N − 1, (1.4) with the conditions v 0 =−α(u 0 ), u N = β(v N ), u j (0) = u j0 , j = 1,N,andv j (0) = v j0 , j = 0,N − 1. For a unitary writing, we take j = 1,N − 1 in both equations of the above system and then the extreme conditions become v 0 =−α(u 1 )andu N = β(v N−1 ) (they do not show u 0 and v N ). By passing N → N + 1 and taking different steps h j , we obtain the system (S) in u j , v j , j = 1,N with v 0 =−α(u 1 ), u N+1 = β(v N )(α = α, β = β), and H = R . In this way the study of the partial differential system (S) reduces to the study of the ordinary differential system ( S)(withH = R and h j = h,forall j). The solution u j , v j depends on h and it seems that u(t,x) = u j (t)ϕ j (x), v(t,x) = v j (t)ϕ j (x) approximate the solution u, v of the system (S). We will not study here the convergence of the solution u, v to u, v, but we will investigate the well-posedness of the discrete problem (S)+(EC)+ ( ID). We will also study the discrete system that corresponds to (S)forx ∈ (0,∞) with the boundary condition v(t,0) =−α u(t,0) , t>0, (BC) 1 and initial data u(0,x) = u 0 (x), v(0, x) = v 0 (x), x>0. (IC) 1 More precisely we will investigate the infinite discrete hyperbolic system du n dt + v n − v n−1 h + A u n f n , dv n dt + u n+1 − u n h + B v n g n , n = 1,2, , t ∈ [0,T], in H, ( S) with the extreme condition v 0 (t) =−α u 1 (t) , t ∈ [0,T], ( EC) and initial data u n (0) = u n0 , v n (0) = v n0 , n = 1,2, ( ID) Although the proposed problems appeared by discretization of the problem (S)+ (BC)+(IC) and the corresponding one for x ∈ (0,∞), our problems also cover some nonlinear differential systems in Hilbert spaces. 4 Existence for a class of discrete hyperbolic problems Forotherclassesofdifference and differential equations in abstract spaces we refer the reader to [1, 2, 10]. In Section 2 we recall some definitions and results from the theory of maximal mono- tone operators that we need to prove our results. In Sections 3 and 4 we study the prob- lems ( S)+(EC)+(ID)and( S)+( EC)+( ID). 2. Notations and preliminaries Let H be a real Hilbert space with the scalar product ·,· and the associated norm ·. We denote by → and the strong and weak convergence in H, respectively. For a multivalued operator A : H → H we denote by D(A) ={x ∈ H; A(x) =∅}its domain and by R(A) =∪{A(x); x ∈ D(A)} its range.TheoperatorA is identified with its graph G(A) ={[x, y] ∈ H × H; x ∈ D(A), y ∈ R(A)}⊂H × H. We use for A the notation A : D(A) ⊂ H → H.IfA,B ⊂ H → H and λ ∈ R then λA = [x, λy]; y ∈ A(x) , D( λA) = D(A), A + B = [x, y + z]; y ∈ A(x), z ∈ A(x) , D( A + B) = D(A) ∩ D(B). (2.1) The operator A : D(A) ⊂ H → H is monotone if for all x 1 ,x 2 ∈ D(A)andy 1 ∈ A(x 1 ), y 2 ∈ A(x 2 )wehavey 1 − y 2 ,x 1 − x 2 ≥0. An operator A : H → H single-valued and everywhere defined is hemicontinuous if for all x, y ∈ H we have A(x + ty) A(x), as t → 0. The operator A : D(A) ⊂ H → H is demi- continuous if it is strongly weakly continuous, that is, if ([x n , y n ]) n ⊂ A with x n → x,as n →∞and y n y,asn →∞,then[x, y] ∈ A. A demicontinuous operator is also hemicontinuous. The operator A : D(A) ⊂ H → H is maximal monotone if it is maximal in the set of all monotone operators, that is, A is monotone and, as subset of H × H, it is not properly contained in any other monotone subset of H × H. The monotone operator A : D(A) ⊂ H → H is maximal monotone if and only if for any λ>0 (equivalently for some λ>0), R(I + λA) = H. If A : H → H is e verywhere defined, single-valued, monotone, and hemicontinuous, then it is maximal monotone. If A : D(A) ⊂ H → H is maximal monotone and B : H → H is everywhere defined, single-valued, monotone, and hemicontinuous, then A + B is maximal monotone. For a maximal monotone operator A : D(A) ⊂ H → H,theoperators J λ = (I + λA) −1 : H −→ H, λ>0, A λ = 1 λ I − J λ : H −→ H, λ>0, (2.2) are the resolvent and the Yosida approximation of A. For an operator A : D(A) ⊂ H → H, f :(0,∞) → H and u 0 ∈ H, we consider the Cauchy problem du dt (t)+A u(t) f (t), t>0, u(0) = u 0 . (CP) Rodica Luca 5 The function u ∈ C([0,T];H)isstrong s olution for the problem (CP)ifu is absolutely continuous on every compact of (0, T), u(t) ∈ D(A), for a.a. t ∈ (0,T), u(0) = u 0 ,andu satisfies (CP) 1 for a.a. t ∈ (0,T). The function u ∈ C([0,T];H)isweak solution for the problem (CP) if there exist (u n ) n ⊂ W 1,∞ (0,T;H)and(f n ) n ⊂ L 1 (0,T;H)suchthat du n dt (t)+A u n (t) f n (t), for a.a. t ∈ (0,T), n = 1,2, , (2.3) u n → u,asn →∞,inC([0,T];H), u(0) = u 0 ,and f n → f ,asn →∞,inL 1 (0,T;H). For other properties of the maximal monotone operators and for the main results of existence, uniqueness of the strong and weak solutions for the nonlinear evolution equations in Hilbert spaces, we refer the reader to [3, 6, 13]. 3. The problem ( S)+(EC)+(ID) The assumptions we will use in this section are the following. (H1) The operators A : D(A) ⊂ H → H, B : D(B) ⊂ H → H are maximal monotone, possibly multivalued, with D(A) =∅, D(B) =∅. (H2) The oper ators α,β : H → H are single-valued and maximal monotone. (H3) The constants h j > 0, j = 1,N. We will write our problem as a Cauchy problem in a certain Hilbert space, for we con- sider the Hilbert space X = H 2N ={(u 1 ,u 2 , ,u N ,v 1 ,v 2 , ,v N ) T ; u j ,v j ∈ H, j = 1, N} with the scalar product u 1 , ,u N ,v 1 , ,v N T , u 1 , ,u N ,v 1 , ,v N T X = N j=1 h j u j ,u j + v j ,v j (3.1) and the corresponding norm · X . We introduce the operator Ꮽ : D(Ꮽ) ⊂ X → X, Ꮽ u 1 ,u 2 , ,u N ,v 1 ,v 2 , ,v N T = v 1 + α u 1 h 1 , v 2 − v 1 h 2 , , v N − v N−1 h N , u 2 − u 1 h 1 , u 3 − u 2 h 2 , , β v N − u N h N T . (3.2) Because D(α) = D(β) = H,wededucethatD(Ꮽ) = X. We also define the operator Ꮾ : D(Ꮾ) ⊂ X → X, D(Ꮾ) = D(A) N × D(B) N , Ꮾ u 1 ,u 2 , ,u N ,v 1 ,v 2 , ,v N T = γ 1 ,γ 2 , ,γ N ,δ 1 ,δ 2 , ,δ N T ; γ i ∈ A u i , δ i ∈ B v i , i = 1, N . (3.3) 6 Existence for a class of discrete hyperbolic problems Using t he operators Ꮽ and Ꮾ, o ur problem can be equivalently expressed as the fol- lowing Cauchy problem in the space X dU dt (t)+Ꮽ U(t) + Ꮾ U(t) F(t), U(0) = U 0 ,(P) where U = (u 1 ,u 2 , ,u N ,v 1 , ,v N ) T , U 0 = (u 10 ,u 20 , ,u N0 ,v 10 , ,v N0 ) T , F = ( f 1 , f 2 , , f N ,g 1 , ,g N ) T . Lemma 3.1. If the assumptions (H2) and (H3) hold, then the operator Ꮽ is monotone and demicontinuous; so it is maximal monotone. Proof. The operator Ꮽ is defined on X and it is single-valued. Ꮽ is monotone, because Ꮽ(U) − Ꮽ(U),U − U X = N j=1 h j v j − v j−1 − v j + v j−1 h j ,u j − u j + N j=1 h j u j+1 − u j − u j+1 + u j h j ,v j − v j = v 1 + α u 1 − v 1 − α u 1 ,u 1 − u 1 + N j=2 v j − v j ,u j − u j − v j−1 − v j−1 ,u j − u j + N−1 j=1 u j+1 − u j+1 ,v j − v j − u j − u j ,v j − v j + β v N − u N − β v N + u N ,v N − v N = v 1 − v 1 ,u 1 − u 1 + α u 1 − α u 1 ,u 1 − u 1 − u 1 − u 1 ,v 1 − v 1 + v N − v N ,u N − u N + β v N − β v N ,v N − v N − u N − u N ,v N − v N = α u 1 − α u 1 ,u 1 − u 1 + β v N − β v N ,v N − v N ≥ 0, (3.4) with v 0 =−α(u 1 ), v 0 =−α(u 1 ), u N+1 = β(v N ), u N+1 = β(v N+1 ), for all U = (u 1 ,u 2 , ,u N ,v 1 , ,v N ) T , U = (u 1 ,u 2 , ,u N ,v 1 , ,v N ) T ∈ X. The operator Ꮽ is also demicontinuous, that is, if U n → U 0 and Ꮽ(U n ) V 0 ,then V 0 = Ꮽ(U 0 ). Indeed, let U n = ( u n 1 ,u n 2 , ,u n N ,v n 1 , ,v n N ), U 0 = ( u 0 1 ,u 0 2 , ,u 0 N ,v 0 1 , ,v 0 N ), U n → U 0 and Ꮽ(U n ) = ((v n 1 + α(u n 1 ))/h 1 ,(v n 2 − v n 1 )/h 2 , ,(v n N − v n N −1 )/h N ,(u n 2 − u n 1 )/ h 1 , ,(β(v n N ) − u n N )/h N ) T , V 0 = (x 0 1 ,x 0 2 , ,x 0 N , y 0 1 , , y 0 N ) T , Ꮽ(U n ) V 0 . Rodica Luca 7 From U n → U 0 we deduce that u n j −→ u 0 j , v n j −→ v 0 j ,forn −→ ∞ ,∀ j = 1,N. (3.5) Because Ꮽ(U n ) V 0 ,wegetᏭ(U n ),Y X →V 0 ,Y X ,forallY ∈ X, Y = (α 1 ,α 2 , ,α N , β 1 , ,β N ) T , that is, v n 1 + α u n 1 ,α 1 −→ h 1 x 0 1 ,α 1 , v n j − v n j −1 ,α j −→ h j x 0 j ,α j , j = 2,N, u n j − u n j −1 ,β j−1 −→ h j−1 y 0 j −1 ,β j−1 , j = 2,N, β v n N − u n N ,β N −→ h N y 0 N ,β N ,inH, =⇒ v n 1 + α u n 1 h 1 x 0 1 , (3.6) v n j − v n j −1 h j x 0 j , j = 2,N, u n j − u n j −1 h j−1 y 0 j −1 , j = 2,N, (3.7) β v n N − u n N h N y 0 N ,inH,asn −→ ∞ . (3.8) From the relations (3.5)and(3.7)weobtainx 0 j = (v 0 j − v 0 j −1 )/h j , j = 2,N, y 0 j −1 = (u 0 j − u 0 j −1 )/h j−1 , j = 2,N. Because α and β are demicontinuous, by (3.5), (3.6 ), and (3.8)we deduce h 1 x 0 1 − v 0 1 = α u 0 1 , h N y 0 N + u 0 N = β v 0 N =⇒ x 0 1 = v 0 1 + α u 0 1 h 1 , y 0 N = β v 0 N − u 0 N h N . (3.9) Therefore V 0 = Ꮽ(U 0 ). Hence the operator Ꮽ is demicontinuous (so it is also hemicon- tinuous) and, by [6, Proposition 2.4] we deduce that it is maximal monotone. Lemma 3.2. If the assumptions (H1) and (H3) hold, then the operator Ꮾ is maximal mono- tone in X. Proof. The operator Ꮾ is evidently monotone Z − Z,U − U X = N j=1 h j γ j − γ j ,u j − u j + δ j − δ j ,v j − v j ≥ 0, (3.10) for all U = (u 1 ,u 2 , ,u N ,v 1 , ,v N ) T , U = (u 1 ,u 2 , ,u N ,v 1 , ,v N ) T ∈ D(Ꮾ), Z ∈ Ꮾ(U), Z ∈ Ꮾ(U), for all γ j ∈ A(u j ), γ j ∈ A(u j ), δ j ∈ B( v j ), δ j ∈ B(v j ), j = 1,N. It is also maximal monotone in X. Indeed, by [6, Proposition 2.2] it is sufficient (and necessary) to show that for λ>0, R(I + λᏮ) = X ⇔ for all Y ∈ X, Y = (x 1 ,x 2 , ,x N , y 1 , , y N ) T there exists U ∈ X, U = (u 1 ,u 2 , ,u N ,v 1 , ,v N ) T such that U + λᏮ(U) Y. (3.11) 8 Existence for a class of discrete hyperbolic problems The last relation gives us u j + λγ j = x j , j = 1,N, v j + λδ j = y j , j = 1,N, =⇒ u j = I + λA −1 x j = J A λ x j , j = 1,N, v j = I + λB −1 y j = J B λ y j , j = 1,N, (3.12) where γ j ∈ A(u j ), δ j ∈ B(v j ) j = 1,N,andJ A λ and J B λ are the resolvents of A and B,re- spectively (A and B are maximal monotone). Then U = (u 1 , ,u N ,v 1 , ,v N ) T ,whereu j and v j , j = 1,N, defined above, satisfy our condition (3.11). We give now the main result for our initial problem (S)+(EC)+(ID). Theorem 3.3. Assume that the assumpt ions (H1)–(H3) hold. If u j0 ∈ D(A),forallj = 1,N, v j0 ∈ D(B),forallj = 1,N,and f j ,g j ∈ W 1,1 (0,T;H), j = 1, N, then there exist unique functions u j and v j ∈ W 1,∞ (0,T;H), j = 1,N, u j (t) ∈ D(A), v j (t) ∈ D(B),forall j = 1,N,forallt ∈ [0,T], which verify the system (S)foreveryt ∈ [0,T), the condition ( EC)foreveryt ∈ [0,T), and the initial data (ID). Moreover u j and v j , j = 1, N, are everywhere differentiable from right in the topology of H and d + u j dt = f j − A u j − v j − v j−1 h j 0 , j = 1,N, d + v j dt = g j − B v j − u j+1 − u j h j 0 , j = 1,N, ∀t ∈ [0,T), (3.13) with v 0 (t) =−α(u 1 (t)), u N+1 (t) = β(v N (t)), ∀t ∈ [0,T). Proof. Because the operator Ꮾ is maximal monotone in X and Ꮽ is sing le-valued, with D( Ꮽ) = X, monotone, and hemicontinuous, by [3, Corollary 1.3, Chapter II] we deduce that Ꮽ + Ꮾ : D(Ꮾ) ⊂ X → X is maximal monotone. By [3, Theorem 2.2, Corollary 2.1, Chapter III] we deduce that, for U 0 ∈ D(Ꮾ)andF ∈ W 1,1 (0,T;X), the problem (P)has a unique solution U = (u 1 ,u 2 , ,u N ,v 1 , ,v N ) T ∈ W 1,∞ (0,T;X), U(t) ∈ D(Ꮾ), for all t ∈ [0,T). We consider (P) 1 in the interval [0,T + ε], ε>0, (by extending correspondingly the functions f j and g j , j = 1,N)andwegetU(T) ∈ D(Ꮾ). The solution U is everywhere differentiable from right and d + U dt (t) = F(t) − Ꮽ U(t) − Ꮾ U(t) 0 , ∀t ∈ [0,T), (3.14) that is, the relations from theorem are verified. In addition we have d + U dt (t) X ≤ F(0)− Ꮽ U 0 − Ꮾ U 0 0 X + t 0 dF ds (s) X ds, ∀t ∈ [0,T). (3.15) Rodica Luca 9 If U and V are the solutions of ( P) corresponding to (U 0 ,F), (V 0 ,G) ∈ D(Ꮾ) × W 1,1 (0, T;X), then U(t) − V(t) X ≤ U 0 − V 0 X + t 0 F(s) − G(s) X ds, ∀t ∈ [0,T]. (3.16) Remark 3.4. If U 0 ∈ D(Ꮾ) = D(A) N × D(B) N and F ∈ L 1 (0,T;X), then, by [3,Corollary 2.2, Chapter III], the problem ( P) ⇔ (S)+(EC)+(ID) has a unique weak solution U ∈ C([0,T];X), that is, there exist (F n ) n ⊂ W 1,1 (0,T;X), F n → F,asn →∞,inL 1 (0,T;X)and (U n ) n ⊂ W 1,∞ (0,T;X), U n (0) = U 0 , U n → U,asn →∞in C([0,T];X), st rong solutions for the problems dU n dt (t)+(Ꮽ + Ꮾ) U n (t) F n (t), for a.a. t ∈ (0,T), n = 1,2, (3.17) 4. The problem ( S)+( EC)+( ID) We present the assumptions that we will use in this section as follows. ( H1) The operators A : D(A) ⊂ H → H and B : D(B) ⊂ H → H are maximal mono- tone, 0 ∈ A(0), 0 ∈ B(0), and there exist a 1 ,a 2 > 0suchthat γ≤a 1 u, ∀u ∈ D(A), ∀γ ∈ A(u); δ≤a 2 u, ∀u ∈ D(B), ∀δ ∈ B(u). (4.1) ( H2) The operator α : H → H is single-valued and maximal monotone. ( H3) The constant h>0. We consider the space Y = l 2 h (H) × l 2 h (H), where l 2 h (H) ={(u n ) n ⊂ H; ∞ n=1 u n 2 < ∞} (= l 2 (H)), with the scalar product u n n , v n n , u n n , v n n Y = u n n , u n n l 2 h (H) + v n n , v n n l 2 h (H) = ∞ n=1 h u n ,u n + ∞ n=1 h v n ,v n . (4.2) We define the operator Ꮽ : Y → Y, Ꮽ u n n , v n n = v n − v n−1 h n , u n+1 − u n h n ,withv 0 =−α u 1 , (4.3) and the operator Ꮾ : D( Ꮾ) ⊂ Y → Y, Ꮾ u n n , v n n = γ n n , δ n n ∈ Y, γ n ∈ A u n , δ n ∈ B v n , ∀n ≥ 1 , (4.4) with D( Ꮾ) ={((u n ) n ,(v n ) n ) ∈ Y; u n ∈ D(A), v n ∈ D(B), ∀n ≥ 1}. 10 Existence for a class of discrete hyperbolic problems Lemma 4.1. If the assumptions ( H2)and( H3) hold, then the ope rator Ꮽ is monotone and demicontinuous in Y. Proof. First we observe that Ꮽ is well-defined in Y .If((u n ) n ,(v n ) n ) ∈ Y,then Ꮽ((u n ) n , (v n ) n ) ∈ Y,andD( Ꮽ) = Y. The operator Ꮽ is monotone, because Ꮽ u n n , v n n − Ꮽ u n n , v n n , u n n , v n n − u n n , v n n Y = v n − v n−1 h n − v n − v n−1 h n , u n − u n n l 2 h (H) + u n+1 − u n h n − u n+1 − u n h n , v n − v n n l 2 h (H) = α u 1 − α u 1 ,u 1 − u 1 ≥ 0. (4.5) Next we prove that Ꮽ is demicontinuous, that is, if u j n n , v j n n −→ u 0 n n , v 0 n n ,forj −→ ∞ in Y, (4.6) Ꮽ u j n n , v j n n x n n , y n n ,forj −→ ∞ in Y, (4.7) then ((x n ) n ,(y n ) n ) = Ꮽ((u 0 n ) n ,(v 0 n ) n ). From (4.6)wededuce h u j n n − u 0 n n 2 l 2 (H) + h v j n n − v 0 n n 2 l 2 (H) −→ 0, for j −→∞=⇒ u j n n − u 0 n n l 2 (H) −→ 0, for j −→ ∞ v j n n − v 0 n n l 2 (H) −→ 0, for j −→ ∞ =⇒ ∞ n=1 u j n − u 0 n 2 −→ 0, for j −→ ∞ ∞ n=1 v j n − v 0 n 2 −→ 0, for j −→ ∞ =⇒ u j n −→ u 0 n ,forj −→ ∞ , ∀n, v j n −→ v 0 n ,forj −→ ∞ , ∀n. (4.8) Then by (4.7)wehave Ꮽ u j n n , v j n n , α n n , β n n Y −→ x n n , y n n , α n n , β n n Y ,asj −→ ∞ , ∀ α n n , β n n ∈ Y, =⇒ v j n − v j n −1 h n , u j n+1 − u j n h n , α n n , β n n Y , [...]... 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Nonlinear Partial Differential Equations and Related Analysis, Contemporary Mathematics, vol 371, American Mathematical Society, Rhode Island, 2005, pp 231–263 Rodica Luca: Department of Mathematics, “Gheorghe Asachi” Technical University of Iasi, ¸ Boulevard Carol I number 11, Iasi 700506, Romania ¸ E-mail address: rluca@math.tuiasi.ro ... for all t ∈ [0,T), and the initial data (ID) Proof By Lemmas 4.1 and 4.2 we have D(Ꮽ + Ꮾ) = D(Ꮾ) and, by [3, Corollary 1.3, Chapter II], Ꮽ + Ꮾ is maximal monotone Using again [3, Theorem 2.2, Chapter III] we obtain the conclusion of the theorem In addition un and vn are everywhere differentiable from the right on [0, T) and, by extended fn and gn on [0, T + ε] with ε > 0, we have V (t) ∈ D(Ꮾ), for all... 0 Therefore (((vn − vn−1 )/h)n ,((u0 − u0 )/h)n ) = ((xn )n ,(yn )n ), (v0 = −α(u0 )) We deduce n+1 1 n that Ꮽ is demicontinuous, so it is maximal monotone Lemma 4.2 If the assumptions (H1) and (H3) hold, then the operator Ꮾ is maximal monotone in Y Proof We suppose without loss of generality (for an easy writing) that A and B are singlevalued The operator Ꮾ under the assumptions of this lemma is well-defined... [8] A Halanay and V R˘ svan, Frequency domain conditions for forced oscillations in difference a ´ ´ systems, Revue Roumaine des Sciences Techniques S´ rie Electrotechnique et Energ´ tique 24 e e (1979), no 1, 141–148 14 Existence for a class of discrete hyperbolic problems [9] V Iftimie, Sur un probl`me non lin´aire pour un syst`me hyperbolique, Revue Roumaine de e e e Math´ matiques Pures et Appliqu´... well-defined in Y Indeed, for ((un )n ,(vn )n ) ∈ D(Ꮾ) ⇔ ((un )n ,(vn )n ) ∈ Y , un ∈ D (A) , vn ∈ D(B), for all n ≥ 1, we have Ꮾ((un )n ,(vn )n ) ∈ Y , that is, (A( un ))n ,(B(vn ))n ∈ l2 (H) By (H1) we have ∞ A un 2 ∞ n =1 2 = a2 1 un a2 2 2 = a2 2 2 vn n l2 (H) n =1 ∞ B vn n =1 a2 u n 1 ≤ 2 n =1 < ∞, (4.12) ∞ ≤ 2 n l2 (H) vn < ∞ 12 Existence for a class of discrete hyperbolic problems The operator Ꮾ is monotone . initial-boundary value problems for linear hyperbolic partial differential equations, Journal of Mathematical Anal- ysis and Applications 24 (1968), no. 2, 372–387. [8] A. Halanay and V. R ˘ asvan,. Editura Academiei, Bucharest, 1988. [14] A. Rousseau, R. Temam, and J. Tribbia, Boundary conditions for an ocean related system with a small parameter, Nonlinear Partial Differential Equations and. systems,Journalof Mathematical Analysis and Applications 83 (1981), no. 2, 470–485. [13] , Nonlinear Evolution Equations and Applications, Mathematics and Its Applications (East European Series),