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 Email 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 wellposedness 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 singlevalued 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, singlevalued, monotone, and hemicontinuous,
then it is maximal monotone.
If A : D(A) H H is maximal monotone and B : H H is everywhere defined,
singlevalued, 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 singlevalued. 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 singlevalued, 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 welldefined 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.