Existence for a class of discrete hyperbolic problems

Advances in Difference Equations, Aug 2006

We investigate the existence and uniqueness of solutions to a class of discrete hyperbolic systems with some nonlinear extreme conditions and initial data, in a real Hilbert space.

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Existence for a class of discrete hyperbolic problems

0 Rodica Luca: Department of Mathematics, Gheorghe Asachi Technical University of Ias i , Boulevard Carol I number 11, Ias i 700506, Romania E-mail address: We investigate the existence and uniqueness of solutions to a class of discrete hyperbolic systems with some nonlinear extreme conditions and initial data, in a real Hilbert space. - with the extreme conditions and the initial data 1. Introduction v j (t) v j1(t) u j+1(t) u j (t) + A u j (t) j = 1, N , t [0, T], in H, uN+1(t) = vN (t) , t [0, T], u j (0) = u j0, v j (0) = v j0, j = 1, N , where N N, h j > 0, j = 1, N , and , , A, B are operators in H, which satisfy some assumptions. (t, x) + A u(t, x) (t, x) + B v(t, x) 0 < x < 1, t > 0, in R, with the boundary conditions and the initial data u(0, x) = u0(x), v(0, x) = v0(x), 0 < x < 1. 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, Morosanu, and Luca, in [4, 5, 9, 11, 12]. Using an idea from [14] we discretize 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 conditions (BC) by + A u(t, x) + B v(t, x) f (t, x), x h, (N + 1)h , g(t, x), x (0, N h), t > 0, We look for u and v of the form u(t, x) = where We write f , g, u0, v0 as j=0 N j=0 f (t, x) = f j (t)j (x), g(t, x) = g j (t)j (x), u0(x) = u j0 j (x), v0(x) = v j0 j (x), j=0 Then for u j and v j we obtain the system j = 1, N , j = 0, N 1, with the conditions v0 = (u0), uN = (vN ), u j (0) = u j0, j = 1, N , and v 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 v0 = (u1) and uN = (vN1) (they do not show u0 and vN ). 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 v0 = (u1), uN+1 = (vN ) ( = , = ), and H = R. In this way the study of the partial differential system (S) reduces to the study of the ordinary differential system (S) (with H = R and h j = h, for all j). The solution u j , v j depends on h and it seems that u(t, x) = u j (t) j (x), v(t, x) = v j (t) 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) for x (0, ) with the boundary condition and initial data u(0, x) = u0(x), v(0, x) = v0(x), More precisely we will investigate the infinite discrete hyperbolic system n = 1, 2, . . . , t [0, T], in H, with the extreme condition and initial data v0(t) = u1(t) , t [0, T], un(0) = un0, vn(0) = vn0, n = 1, 2, . . . . 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. For other classes of difference 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 monotone operators that we need to prove our results. In Sections 3 and 4 we study the problems (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. The operator A 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. If A, B H H and R then A + B = [x, y + z]; y A(x), z A(x) , D(A + B) = D(A) D(B). The operator A : D(A) H H is monotone if for all x1, x2 D(A) and y1 A(x1), y2 A(x2) we have y1 y2, x1 x2 0. An operator A : H H single-valued and everywhere defined is hemicontinuous if for all x, y H we have A(x + t y) A(x), as t 0. The operator A : D(A) H H is demicontinuous if it is strongly weakly continuous, that is, if ([xn, yn])n A with xn x, as n and yn y, as n , 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 > 0 (equivalently for some > 0), R(I + A) = H. If A : H H is everywhere 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, the operators J = (I + A)1 : H H, > 0, f (t), t > 0, u(0) = u0. The function u C([0, T]; H) is strong solution for the problem (CP) if u is absolutely continuous on every compact of (0, T), u(t) D(A), for a.a. t (0, T), u(0) = u0, and u satisfies (CP)1 for a.a. t (0, T). The function u C([0, T]; H) is weak solution for the problem (CP) if there exist (un)n W 1,(0, T; H) and ( fn)n L1(0, T; H) such that fn(t), for a.a. t (0, T), n = 1, 2, . . . , un u, as n , in C([0, T]; H), u(0) = u0, and fn f , as n , in L1(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) u1, . . . , uN , v1, . . . , vN T , u1, . . . , uN , v1, . . . , vN T X = j=1 and the corresponding norm We introduce the operator u1, u2, . . . , uN , v1, v2, . . . , vN Because D() = D() = H, we deduce that D( ) = X. We also define the operator : D( ) X X, D( ) = D(A)N D(B)N , u1, u2, . . . , uN , v1, v2, . . . , vN U(0) = U0, Lemma 3.1. If the assumptions (H2) and (H3) hold, then the operator demicontinuous; so it is maximal monotone. is monotone and Proof. The operator is defined on X and it is single-valued. is monotone, because j=1 j=1 j=2 N1 j=1 (U) (U), U U X , vj v j = v1 + u1 v1 u1 , u1 u1 vj v j, uj uj vj1 v j1, uj uj uj+1 uj+1, vj v j uj uj, vj v j + vN uN vN + uN , vN vN = v1 v1, u1 u1 + u1 u1 , u1 u1 u1 u1, v1 v1 + vN vN , uN uN + vN vN , vN vN uN uN , vN vN = u1 u1 , u1 u1 + vN vN , vN vN 0, with v0 = (u1), v0 = (u1), uN+1 = (vN ), uN+1 = (vN+1), for all U = (u1, u2, . . . , uN , v1, . . . , vN )T , U = (u1, u2, . . . , uN , v1, . . . , vN )T X. The operator is also demicontinuous, that is, if Un U0 and (Un) V 0, then V 0 = (U0). Indeed, let Un = (u1n, u2n, . . . , unN , v1n, . . . , vNn ), U0 = (u10, u20, . . . , u0N , v10, . . . , vN0 ), Un U0 and (Un) = ((v1n + (u1n))/h1, (v2n v1n)/h2, . . . , (vNn vNn1)/hN , (u2n u1n)/ h1, . . . , ((vNn ) unN )/hN )T , V 0 = (x10, x20, . . . , xN0 , y10, . . . , yN0 )T , (Un) V 0. unj u0j , vnj v0j, for n , j = 1, N. (Un), Y X V 0, Y X , for all Y X, Y = (1, 2, . . . , N , From Un U0 we deduce that v1n + u1n , 1 h1 x10, 1 , vnj vnj1, j hj x0j, j , j = 2, N, unj unj1, j1 hj1 y0j1, j1 , j = 2, N, vNn unN , N hN yN0 , N , in H, vnj vnj1 unj unj1 hjx0j, j = 2, N, hj1 y0j1, j = 2, N, hN yN0 , in H, as n . hN yN0 + u0N = vN0 = x1 = 0 Therefore V 0 = (U0). Hence the operator is demicontinuous (so it is also hemicontinuous) 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 is maximal monotone in X. Proof. The operator is evidently monotone j=1 Z Z, U U X = hj j j, uj uj + j j, vj v j 0, for all U = (u1, u2, . . . , uN , v1, . . . , vN )T , U = (u1, u2, . . . , uN , v1, . . . , vN )T D( ), Z (U), Z (U), for all j A(uj), j A(uj), j B(vj), 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 > 0, R(I + ) = X for all Y X, Y = (x1, x2, . . . , xN , y1, . . . , yN )T there exists U X, U = (u1, u2, . . . , uN , v1, . . . , vN )T such that The last relation gives us j = 1, N , u j = I + A 1 x j = JA x j , j = 1, N , = v j = I + B 1 y j = JB y j , j = 1, N , j = 1, N , where j A(u j ), j B(v j ) j = 1, N , and JA and JB are the resolvents of A and B, respectively (A and B are maximal monotone). Then U = (u1, . . . , uN , v1, . . . , vN )T , where u 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 assumptions (H1)(H3) hold. If u j0 D(A), for all j = 1, N , v j0 D(B), for all j = 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), for all j = 1, N , for all t [0, T], which verify the system (S) for every t [0, T), the condition (EC) for every t [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 dt = = g j B v j u j+1 u j 0 v j v j1 0 f j A u j j = 1, N , j = 1, N , t [0, T), Proof. Because the operator is maximal monotone in X and is single-valued, with D( ) = X, monotone, and hemicontinuous, by [3, Corollary 1.3, Chapter II] we deduce that + : D( ) X X is maximal monotone. By [3, Theorem 2.2, Corollary 2.1, Chapter III] we deduce that, for U0 D( ) and F W 1,1(0, T; X), the problem (P) has a unique solution U = (u1, u2, . . . , uN , v1, . . . , vN )T W 1,(0, T; X), U(t) D( ), for all t [0, T). We consider (P)1 in the interval [0, T + ], > 0, (by extending correspondingly the functions f j and g j , j = 1, N ) and we get U(T) D( ). The solution U is everywhere differentiable from right and U(t) t [0, T), that is, the relations from theorem are verified. In addition we have F(0) U0 If U and V are the solutions of (P) corresponding to (U0, F), (V0, G) D( ) W1,1(0, T; X), then U(t) V (t) X U0 V0 X + F(s) G(s) X ds, t [0, T]. Remark 3.4. If U0 D( ) = D(A)N D(B)N and F L1(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 (Fn)n W1,1(0, T; X), Fn F, as n , in L1(0, T; X) and (Un)n W1,(0, T; X), Un(0) = U0, Un U, as n in C([0, T]; X), strong solutions for the problems Fn(t), for a.a. t (0, T), n = 1, 2, . . . . 4. The problem (S) + (EC) + (ID) a2 u , u D(B), B(u). (4.1) un n, vn n , un n, vn n Y = n=1 n=1 We define the operator : Y Y , and the operator : D( ) Y Y , un n, vn n = un n, vn n = n n, n n Y , n A un , n B vn , n 1 , un n, vn n un n, vn n , un n, vn n = u1 u1 , u1 u1 0. Next we prove that is demicontinuous, that is, if unj n, vnj n u0n n, vn0 n , for j in Y , xn n, yn n , for j in Y , then ((xn)n, (yn)n) = ((u0n)n, (vn0)n). From (4.6) we deduce unj n u0n n l2(H) 0, for j vnj n vn0 n l2(H) 0, for j unj u0n 2 0, for j vnj vn0 2 0, for j unj u0n, for j , n, Lemma 4.1. If the assumptions (H2) and (H3) hold, then the operator demicontinuous in Y . is monotone and Proof. First we observe that (vn)n) Y , and D( ) = Y . The operator is monotone, because n=1 Then by (4.7) we have n=1 n=1 n=1 h yn, n , as j , with v0j = u1j , j 1, n=1 n=1 n=1 a22 vn 2 = a22 vn n l22(H) < . u1j = v1j + u1j v1j hx1 v10, as j (v1j v10 as j , by (4.8) with n = 1). Now since is a demicontinuous operator, and moreover u1j u10, as j (by (4.8) with n = 1), we have Therefore (((vn0 vn01)/h)n, ((u0n+1 u0n)/h)n) = ((xn)n, (yn)n), (v00 = (u10)). We deduce that is demicontinuous, so it is maximal monotone. Lemma 4.2. If the assumptions (H1) and (H3) hold, then the operator is maximal monotone in Y . Proof. We suppose without loss of generality (for an easy writing) that A and B are singlevalued. The operator under the assumptions of this lemma is well-defined in Y . Indeed, for ((un)n, (vn)n) D( ) ((un)n, (vn)n) Y , un D(A), vn D(B), for all n 1, we have ((un)n, (vn)n) Y , that is, (A(un))n, (B(vn))n l2(H). By (H1) we have Existence for a class of discrete hyperbolic problems The operator is monotone Moreover is maximal monotone, that is, > 0, R(I + B) = Y Z = xn n, yn n Y , W = vn0 D(B), for all n 1 with (un0)n, (vn0)n l2(H)) and F W 1,1(0, T; Y ) (( fn)n, (gn)n W 1,1(0, T; l2(H))), then there exists a unique function V = ((un)n, (vn)n) W 1,(0, T; Y ), with V (t) D(B), for all t [0, T], and V verifies the system (S), for all t [0, T), the extreme condition (EC), for all t [0, T), and the initial data (ID). Proof. By Lemmas 4.1 and 4.2 we have D( + ) = D( ) and, by [3, Corollary 1.3, Chapter II], + 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 + ] with > 0, we have V (t) D( ), for all t [0, T]. Remark 4.4. Although the operators and have semblable forms, we cannot establish connections between them, because the spaces, where these operators are defined, are different. Indeed, defining un = vn = 0, for all n N + 1, we have un n, vn n = Therefore Theorem 4.3. The author would like to thank the referees for the careful reading of the manuscript and for their comments which led to this revised version.


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Rodica Luca. Existence for a class of discrete hyperbolic problems, Advances in Difference Equations, 2006, 089260,